Integration techniques
usubstitution
None
(2^(ln x))/x antiderivative example
Finding ∫(2^ln x)/x dx
Discussion and questions for this video
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 I was just looking on the discussion boards on the Khan
 Academy Facebook page, and Bud Denny put up this problem,
 asking for it to be solved.
 And it seems like a problem of general interest.
 If the indefinite integral of 2 to the natural log of x
 over, everything over x, dx.
 And on the message board, Abhi Khanna also put up a solution,
 and it is the correct solution, but I thought this was of
 general interest, so I'll make a quick video on it.
 So the first thing when you see an integral like this, is you
 say, hey, you know, I have this natural log of x up in
 the numerator, and where do I start?
 And the first thing that should maybe pop out at you, is that
 this is the same thing as the integral of one over x times 2
 to the natural log of x, dx.
 And so you have an expression here, or it's kind of part of
 our larger function, and you have its derivative, right?
 We know that the derivative, let me write it over here, we
 know that the derivative with respect to x of the natural
 log of x is equal to 1/x.
 So we have some expression, and we have its derivative, which
 tells us that we can use substitution.
 Sometimes you can do in your head, but this problem, it's
 still not trivial to do in your head.
 So let's make the substitution.
 Let's substitute this right here with a u.
 So let's do that.
 So if you define u, and it doesn't have to be u, it's
 just, that's the convention, it's called usubstitution, it
 could have been ssubstitution for all we care.
 Let's say u is equal to the natural log of x, and then du
 dx, the derivative of u with respect to x, of course
 is equal to 1/x.
 Or, just the differential du, if we just multiply both sides
 by dx is equal to 1 over x dx.
 So let's make our substitution.
 This is our integral.
 So this will be equal to the indefinite integral, or the
 antiderivative, of 2 to the now u, so 2 to the u,
 times 1 over x dx.
 Now what is 1 over x dx?
 That's just du.
 So this term times that term is just our du.
 Let me do it in a different color.
 1 over x times dx is just equal to du.
 That's just equal to that thing, right there.
 Now, this still doesn't look like an easy integral, although
 it's gotten simplified a good bit.
 And to solve, you know, whenever I see the variable
 that I'm integrating against in the exponent, you know,
 we don't have any easy exponent rules here.
 The only thing that I'm familiar with, where I have my
 x or my variable that I'm integrating against in
 my exponent, is the case of e to the x.
 We know that the integral of e to the x, dx, is equal
 to e to the x plus c.
 So if I could somehow turn this into some variation of e to the
 x, maybe, or e to the u, maybe I can make this integral a
 little bit more tractable.
 So let's see.
 How can we redefine this right here?
 Well, 2, 2 is equal to what?
 2 is the same thing as e to the natural log of 2, right?
 The natural log of 2 is the power you have to
 raise e to to get 2.
 So if you raise e to that power, you're, of
 course going to get 2.
 This is actually the definition of really, the natural log.
 You raise e to the natural log of 2, you're going to get 2.
 So let's rewrite this, using this I guess we could call
 this this rewrite or I don't want to call it
 quite a substitution.
 It's just a different way of writing the number 2.
 So this will be equal to, instead of writing the number
 2, I could write e to the natural log of 2.
 And all of that to the u du.
 And now what is this equal to?
 Well, if I take something to an exponent, and then to another
 exponent, this is the same thing as taking my base to the
 product of those exponents.
 So this is equal to, let me switch colors, this is equal
 to the integral of e, to the u, e to the, let
 me write it this way.
 e to the natural log of 2 times u.
 I'm just multiplying these two exponents.
 I raise something to something, then raise it again, we know
 from our exponent rules, it's just a product of
 those two exponents.
 du.
 Now, this is just a constant factor, right here.
 This could be, you know, this could just be some number.
 We could use a calculator to figure out what this is.
 We could set this equal to a.
 But we know in general that the integral, this is pretty
 straightforward, we've now put it in this form.
 The antiderivative of e to the au, du, is just
 1 over a e to the au.
 This comes from this definition up here, and of course plus
 c, and the chain rule.
 If we take the derivative of this, we take the derivative
 of the inside, which is just going to be a.
 We multiply that times the one over a, it cancels
 out, and we're just left with e to the au.
 So this definitely works out.
 So the antiderivative of this thing right here is going to be
 equal to 1 over our a, it's going to be 1 over our constant
 term, 1 over the natural log of 2 times our whole
 expression, e e.
 And I'm going to do something.
 This is just some number times u, so I can write it as
 u times some number.
 And I'm just doing that to put in a form that might help us
 simplify it a little bit.
 So it's e to the u times the natural log of 2, right?
 All I did, is I swapped this order.
 I could have written this as e to the natural
 log of 2 times u.
 If this an a, a times u is the same thing as u times a.
 Plus c.
 So this is our answer, but we have to kind of reverse
 substitute before we can feel satisfied that we've taken
 the antiderivative with respect to x.
 But before I do that, let's see if I can simplify
 this a little bit.
 What is, if I have, just from our natural log properties,
 or logarithms, a times the natural log of b.
 We know this is the same thing as the natural
 log of b to the a.
 Let me draw a line here.
 Right?
 That this becomes the exponent on whatever we're taking
 the natural log of.
 So u, let me write this here, u times the natural log of
 2, is the same thing as the natural log of 2 to the u.
 So we can rewrite our antiderivative as being equal
 to 1 over the natural log of 2, that's just that part here,
 times e to the, this can be rewritten based on this
 logarithm property, as the natural log of 2 to the u, and
 of course we still have our plus c there.
 Now, what is e raised to the natural log of 2 to the u?
 The natural log of 2 to the u is the power that you have to
 raise e to to get to 2 to the u, right?
 By definition!
 So if we raise e to that power, what are we going to get?
 We're going to get 2 to the u.
 So this is going to be equal to 1 over the natural log of 2.
 This simplifies to just 2 to the u.
 I drew it up here.
 The natural log a I could just write in general terms, let
 me do it up here, and maybe I'm beating a dead horse.
 But I can in general write any number a as being equal to
 e to the natural log of a.
 This is the exponent you have to raise e to to get a.
 If you raise e to that, you're going to get a.
 So e to the natural log of 2 to the u, that's just 2 to the u.
 And then I have my plus c, of course.
 And now we can reverse substitute.
 What did we set u equal to?
 We defined u, up here, as equal to the natural log of x.
 So let's just reverse substitute right here.
 And so the answer to our original equation, your answer
 to, let me write it here, because it's satisfying when
 you see it, to this kind of fairly convolutedlooking
 antiderivative problem, 2 to the natural log of x over x dx,
 we now find is equal to, we just replaced u with natural
 log of x, because that was our substitution, and 1 over the
 natural log of 2 times 2 to the natural log of x plus c.
 And we're done.
 This isn't in the denominator, the way I wrote it might
 look a little ambiguous.
 And we're done!
 And that was a pretty neat problem, and so thanks
 to Bud for posting that.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

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At around 4:50 onwards Sal shows that the anti derivative of e^au is (1/a)e^au. even though in reverse (taking the derivative) it seems to make sense I can't really understand how he gets the 1/a when taking the anti derivative.
Try making the following substitution:
(by int[] i mean the integral of whats inside the square brackets)
int[e^ax(dx)]
let u = ax,
then du = a(dx)
then dx = du/a
now substitute this back into the original integral:
int[e^ax(dx)] now equals int[e^u(du/a)]
Therefore:
int[e^ax(dx)] = (1/a)int[e^u(du)]
(by int[] i mean the integral of whats inside the square brackets)
int[e^ax(dx)]
let u = ax,
then du = a(dx)
then dx = du/a
now substitute this back into the original integral:
int[e^ax(dx)] now equals int[e^u(du/a)]
Therefore:
int[e^ax(dx)] = (1/a)int[e^u(du)]
Just wanted to add that there is no way around the words "in reverse." You're looking for the function whose derivative is {the function before you}.
In a nutshell, when integrating, you're working backwards.
You don't sound sure about the derivative. Always a good idea to review the derivative, ask it the question you have(like, "what happens to the constants when I differentiate ce^ax ?"), and return to the problem.
In a nutshell, when integrating, you're working backwards.
You don't sound sure about the derivative. Always a good idea to review the derivative, ask it the question you have(like, "what happens to the constants when I differentiate ce^ax ?"), and return to the problem.
Could we not also say 2^Lnx = (e^Ln2)^Lnx = e^(Ln2*Lnx) = (e^Lnx)^Ln2 = x^Ln2 ?
Then we have the integral of (x^Ln2)/x, which should just be the integral of x^(Ln2  1)dx, which is just a simple power rule (add one to exponent, divide by this new exponent). Might be easier for some people to see it this way, as there is no need for substitution, just Ln algebra instead, which we ended up needing anyways.
Then we have the integral of (x^Ln2)/x, which should just be the integral of x^(Ln2  1)dx, which is just a simple power rule (add one to exponent, divide by this new exponent). Might be easier for some people to see it this way, as there is no need for substitution, just Ln algebra instead, which we ended up needing anyways.
In fact if you have studied properties of logarithms in detail, this is a general rule that:
x^log (y) = y^log (x) [ the base of the logarithm can be anything. ]
which would reduce the number of steps required to solve the integral.
However, I solved it in a similar manner.
x^log (y) = y^log (x) [ the base of the logarithm can be anything. ]
which would reduce the number of steps required to solve the integral.
However, I solved it in a similar manner.
same for me. I must say I was puzzled that Khan took such a difficult road and it made me wondered if I had it wrong with my log vs ln.
Yep, this was how I solved it.
Why is this video in the Multivariable Calculus playlist?
Probably irrelevant by now, but it's been moved to the Usubstitution section because the solution is found using Usubstitution.
Really dumb question I know, but how come 2^(ln x) = x^(ln 2)
Not a dumb question.
The key is knowing that 2 = e^ln 2
Review the logarithm properties if you need to on that because that's really useful.
So then
(e^ln 2)^ln x = (e^ln x)^ln 2
exponent properties.
If that part is confusing I like to recall something like
(2^2)^4 = (2^4)^2 = 2^(4*2) = 256
then last we basically repeat the first step
We know e^ln x = x
So
(e^ln x)^ln 2 = x ^(ln 2)
Hope that helps.
The key is knowing that 2 = e^ln 2
Review the logarithm properties if you need to on that because that's really useful.
So then
(e^ln 2)^ln x = (e^ln x)^ln 2
exponent properties.
If that part is confusing I like to recall something like
(2^2)^4 = (2^4)^2 = 2^(4*2) = 256
then last we basically repeat the first step
We know e^ln x = x
So
(e^ln x)^ln 2 = x ^(ln 2)
Hope that helps.
```2^(ln(x)) = x^(ln(2))
ln(2^(ln(x))) = ln(x^(ln(2)))
ln(x)•ln(2) = ln(2)•ln(x)```
ln(2^(ln(x))) = ln(x^(ln(2)))
ln(x)•ln(2) = ln(2)•ln(x)```
There is a much easier way to do this problem, that I came up with.
We know that d/dx (a^x) = a^x lna. Then we can figure out that d/dx (a^x/lna) = a^x because lna is a constant and it can be taken out, then it gets cancelled.
So, antiderivative of a^x dx = a^x/lna !
Then we just have to put u = ln x. We get 2^u which is in the form a^x
We know that d/dx (a^x) = a^x lna. Then we can figure out that d/dx (a^x/lna) = a^x because lna is a constant and it can be taken out, then it gets cancelled.
So, antiderivative of a^x dx = a^x/lna !
Then we just have to put u = ln x. We get 2^u which is in the form a^x
that's also how i would have done this saves a lot of work if you remember that
@ 3:16 I didnt get how' 2 is the same thing as e to the natural log of 2 ' ?can someone please explain?
e^x and lnx are opposite functions. It would be akin to multiplying by some number and then dividing by that same number. The operations cancel one another out.
e to the natural log of *elephant* = elephant.
e to the natural log of *elephant* = elephant.
Natural log of x means "e to the what power equals x"? In this problem, e^ln(2)=2 out because this is what you're doing: e to the what power equals 2? Well, I don't know, but I'm going to take e to that power. e to the power that makes it equal to 2 is obviously 2.
On the other hand, taking the natural log of e^2 is also equal to 2. So, you are asking, "e to the what power equals e^2?". e squared is equal to e squared. So, that would also be just 2.
On the other hand, taking the natural log of e^2 is also equal to 2. So, you are asking, "e to the what power equals e^2?". e squared is equal to e squared. So, that would also be just 2.
A silly question, but I dont seem to find the answer... is there any difference between u` and du/dx or is it just a simple matter of choice which notation one uses?
That's not a silly question at all. Generally u' and du/dx are simply alternative notations for the same concept, with u' being the notation that originated with Newton and du/dx originating with Leibniz. You need to know both notations because they're used interchangeably.
But there's more to the story. Newton's notation is more compact and often convenient, such as when we state the product rule, (uv)' = uv' + u'v. Leibniz's is more precise because it specifies that the derivative is with respect to the variable x (which is merely assumed in Newton's notation). It's also more powerful because it invites us to think of du and dx as infinitesimal quantities, which is helpful in understanding integration, and provides a bridge to further topics such as multivariable calculus and differential equations. It's been suggested that mathematicians on the European continent outpaced British mathematicians for many decades following the invention of calculus because they used the more flexible du/dx notation on the continent while the British stuck with u' out of loyalty to Newton.
But there's more to the story. Newton's notation is more compact and often convenient, such as when we state the product rule, (uv)' = uv' + u'v. Leibniz's is more precise because it specifies that the derivative is with respect to the variable x (which is merely assumed in Newton's notation). It's also more powerful because it invites us to think of du and dx as infinitesimal quantities, which is helpful in understanding integration, and provides a bridge to further topics such as multivariable calculus and differential equations. It's been suggested that mathematicians on the European continent outpaced British mathematicians for many decades following the invention of calculus because they used the more flexible du/dx notation on the continent while the British stuck with u' out of loyalty to Newton.
I prefere Leibniz, especially for computing derivatives.
Eaxmple, a Khan exercise:
A 13 meter ladder is leaning against a wall when its base begins to slide
away from the wall. By the time the base is `x = 12` meters from the wall,
it is moving at a rate of `dx/dt = 5 m/sec`. At that moment, what is the
vertical velocity of the top of the ladder `dy/dt = (in m/sec.\,)`?
Your answer should be negative since the top of the ladder is falling.
`y = sqrt(13^2  x^2)`
`y = (13^2  x^2)^1/2`
`d/dt(y) = d/dt((13^2  x^2)^1/2)`
`d/dt(y) = 1/2*(13^2  x^2)^1/2*d/dt(13^2  x^2)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*d/dt(13^2  x^2)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2  x^2))`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2)  d/dt(x^2))`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2)  2x*d/dt(x))`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2)  2x*dx/dt)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(0  2x*dx/dt)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(2x*dx/dt)`
`d/dt(y) = 2x/2*sqrt(13^2  x^2)*dx/dt`
`dy/dt = [x/sqrt(13^2  x^2)*dx/dt]`
`x = 12`
`dy/dt = 12/sqrt(13^2  12^2)*dx/dt`
`dy/dt = 12/sqrt(169  144)*dx/dt`
`dy/dt = 12/sqrt(25)*dx/dt`
`dy/dt = 12/5*dx/dt`
`dy/dt = 12/5*dx/dt`
`dx/dt = 5`
`dy/dt = 12/5*5`
`dy/dt = 12*5/5`
`dy/dt = [12]`
f' notation wouid be completely useless for my method.
Eaxmple, a Khan exercise:
A 13 meter ladder is leaning against a wall when its base begins to slide
away from the wall. By the time the base is `x = 12` meters from the wall,
it is moving at a rate of `dx/dt = 5 m/sec`. At that moment, what is the
vertical velocity of the top of the ladder `dy/dt = (in m/sec.\,)`?
Your answer should be negative since the top of the ladder is falling.
`y = sqrt(13^2  x^2)`
`y = (13^2  x^2)^1/2`
`d/dt(y) = d/dt((13^2  x^2)^1/2)`
`d/dt(y) = 1/2*(13^2  x^2)^1/2*d/dt(13^2  x^2)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*d/dt(13^2  x^2)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2  x^2))`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2)  d/dt(x^2))`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2)  2x*d/dt(x))`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(d/dt(13^2)  2x*dx/dt)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(0  2x*dx/dt)`
`d/dt(y) = 1/2*sqrt(13^2  x^2)*(2x*dx/dt)`
`d/dt(y) = 2x/2*sqrt(13^2  x^2)*dx/dt`
`dy/dt = [x/sqrt(13^2  x^2)*dx/dt]`
`x = 12`
`dy/dt = 12/sqrt(13^2  12^2)*dx/dt`
`dy/dt = 12/sqrt(169  144)*dx/dt`
`dy/dt = 12/sqrt(25)*dx/dt`
`dy/dt = 12/5*dx/dt`
`dy/dt = 12/5*dx/dt`
`dx/dt = 5`
`dy/dt = 12/5*5`
`dy/dt = 12*5/5`
`dy/dt = [12]`
f' notation wouid be completely useless for my method.
how to integrate ln(x)^2
you have to use recursive integration by parts(which is product rule in reverse):
⌠ln(x)^2dx = ln(x)•⌠ln(x)dx  ⌠ln(x)/x dx;
⌠ln(x)dx = x•ln(x)  ⌠x/xdx= x•ln(x)  x (+C) //we plug this back in
⌠ln(x)^2dx = ln(x)•(x•ln(x)  x)  ⌠ln(x)/x dx; u=ln(x), u' = 1/x; dx=du/(1/x);
. . . . . . . . =ln(x)•(x•ln(x)  x)  ⌠1/(x/x)du;
. . . . . . . . =ln(x)•(x•ln(x)  x)  u +C
. . . . . . . . =ln(x)•(x•ln(x)  x) ln(x) +C
. . . . . . . . =ln(x)•(x•ln(x)  x  1)+C ; hard, but not as bad as (sec(x))^3.
⌠ln(x)^2dx = ln(x)•⌠ln(x)dx  ⌠ln(x)/x dx;
⌠ln(x)dx = x•ln(x)  ⌠x/xdx= x•ln(x)  x (+C) //we plug this back in
⌠ln(x)^2dx = ln(x)•(x•ln(x)  x)  ⌠ln(x)/x dx; u=ln(x), u' = 1/x; dx=du/(1/x);
. . . . . . . . =ln(x)•(x•ln(x)  x)  ⌠1/(x/x)du;
. . . . . . . . =ln(x)•(x•ln(x)  x)  u +C
. . . . . . . . =ln(x)•(x•ln(x)  x) ln(x) +C
. . . . . . . . =ln(x)•(x•ln(x)  x  1)+C ; hard, but not as bad as (sec(x))^3.
Hi, couldn't you use the formula Integral a^x dx = (a^x/ln a) + c
Yes thats correct. Sal was just explaining the concept without using that formula.
That's the rule I was thought in school as well.
I did this myself, and came up with a different answer...
I converted 2 to e^ln(2) first, so the problem would become
INTEGRAL[e^(ln(2)*ln(x))/x dx] I set u = ln(2)*ln(x) and found du to be ln(2)/x.
So 1/ln(2) * INTEGRAL[e^u dx] = 1/ln(2) * e^u
= e^(ln(2)*ln(x))/ln(2) + C
I'd like to know if this is right as well, and if not  where I went wrong.
Thank you very much :)
I converted 2 to e^ln(2) first, so the problem would become
INTEGRAL[e^(ln(2)*ln(x))/x dx] I set u = ln(2)*ln(x) and found du to be ln(2)/x.
So 1/ln(2) * INTEGRAL[e^u dx] = 1/ln(2) * e^u
= e^(ln(2)*ln(x))/ln(2) + C
I'd like to know if this is right as well, and if not  where I went wrong.
Thank you very much :)
You have the same answer, you just didn't simplify. Remember that
e^[ log (2) ∙ log (x) ]
= e^[log (2^log(x))]
= 2^log (x)
so once simplified your answer is:
[2^log x ] / log 2 + C
Please be sure you know your log and exponent properties very thoroughly, as you will use them extensively in calculus.
e^[ log (2) ∙ log (x) ]
= e^[log (2^log(x))]
= 2^log (x)
so once simplified your answer is:
[2^log x ] / log 2 + C
Please be sure you know your log and exponent properties very thoroughly, as you will use them extensively in calculus.
You could have taken u=2^ln x differentiated and then substituted right?
I'm not sure what you're suggesting, but I suspect it wouldn't work. To make a usubstitution work, the formula has to include both the thing you're choosing as u and the derivative of u (that is, du). Sal is able to do a usubstitution using ln x here because the formula also includes 1/x, the derivative of ln x. We can't do a usubstitution using 2^(ln x) because the formula doesn't contain anything corresponding to the derivative of that expression.
Can you make a video solving ∫ (x² ) arcsin x dx
You need to use integration by parts:
∫u dv = uv  ∫ v du
Let u = arcsin x
Thus du = 1/√(1x²)
Let dv = x²
Thus v = ⅓x³
Thus,
∫ (x² ) arcsin x dx = ⅓ x³ arcsin x − ⅓∫x³/√(1x²) dx
Now we have an integral that we can work with with ordinary u substitution. So, let us integrate just:
∫ x³ / √(1  x²) dx
Let u = 1  x²
Thus, du = 2x dx
Also expressed as ½ du = x dx
And rearranging u = 1  x² we get x² = 1  u
Now let us apply the u sub. Note that x³ = x∙x² as this is used.
∫ x³ / √(1  x²) dx = ∫ ½ (1  u) / √u du
∫ x³ / √(1  x²) dx = ∫ ½ (u  1) / √u du
∫ x³ / √(1  x²) dx = ½ ∫ (u  1) / √u du
Now we split the fraction:
∫ x³ / √(1  x²) dx = ½ ∫ { u/√u} − 1 /√u du}
Now split the integral:
∫ x³ / √(1  x²) dx = ½ ∫( u/√u) du − ½ ∫ 1 /√u du
Simplify the first integral:
∫ x³ / √(1  x²) dx = ½ ∫√u du − ½ ∫ 1 /√u du
To make this easier to follow, express radicals in fractional exponent form:
∫ x³ / √(1  x²) dx = ½ ∫u^(½) du − ½ ∫ u^(−½ du)
Now we finally have something that is easy to integrate:
∫ x³ / √(1  x²) dx = ½(⅔) u^(³⁄₂) − ½(2) u^(½) + C
∫ x³ / √(1  x²) dx = ⅓ u^(³⁄₂) − u^(½) + C
Factor out ⅓√u
∫ x³ / √(1  x²) dx = ⅓(√u) [ u − 3] + C
Rearrange:
∫ x³ / √(1  x²) dx = ⅓) [ u − 3] (√u) + C
Back subbing u = 1  x²
∫ x³ / √(1  x²) dx = ⅓ [1 x² 3] [√(1x²)] + C
∫ x³ / √(1  x²) dx = ⅓ [ x² 2] √(1x²) + C
Factor out 1:
∫ x³ / √(1  x²) dx = −⅓ [ x² + 2] √(1x²) + C
Now we are ready to apply this to original integral:
∫ (x² ) arcsin x dx = ⅓x³ arcsin x − ⅓∫x³/√(1x²) dx
∫ (x² ) arcsin x dx = ⅓x³ arcsin x − (⅓)( −⅓) [ x² + 2] √(1x²) + C
∫ (x² ) arcsin x dx = ⅓x³ arcsin x + (¹⁄₉) [ x² + 2] √(1x²) + C
If you like, to make this a little more simple, factor out ¹⁄₉
∫ (x² ) arcsin x dx = ¹⁄₉ { 3 x³ arcsin x + ( x² + 2) √(1x²)} + C
∫u dv = uv  ∫ v du
Let u = arcsin x
Thus du = 1/√(1x²)
Let dv = x²
Thus v = ⅓x³
Thus,
∫ (x² ) arcsin x dx = ⅓ x³ arcsin x − ⅓∫x³/√(1x²) dx
Now we have an integral that we can work with with ordinary u substitution. So, let us integrate just:
∫ x³ / √(1  x²) dx
Let u = 1  x²
Thus, du = 2x dx
Also expressed as ½ du = x dx
And rearranging u = 1  x² we get x² = 1  u
Now let us apply the u sub. Note that x³ = x∙x² as this is used.
∫ x³ / √(1  x²) dx = ∫ ½ (1  u) / √u du
∫ x³ / √(1  x²) dx = ∫ ½ (u  1) / √u du
∫ x³ / √(1  x²) dx = ½ ∫ (u  1) / √u du
Now we split the fraction:
∫ x³ / √(1  x²) dx = ½ ∫ { u/√u} − 1 /√u du}
Now split the integral:
∫ x³ / √(1  x²) dx = ½ ∫( u/√u) du − ½ ∫ 1 /√u du
Simplify the first integral:
∫ x³ / √(1  x²) dx = ½ ∫√u du − ½ ∫ 1 /√u du
To make this easier to follow, express radicals in fractional exponent form:
∫ x³ / √(1  x²) dx = ½ ∫u^(½) du − ½ ∫ u^(−½ du)
Now we finally have something that is easy to integrate:
∫ x³ / √(1  x²) dx = ½(⅔) u^(³⁄₂) − ½(2) u^(½) + C
∫ x³ / √(1  x²) dx = ⅓ u^(³⁄₂) − u^(½) + C
Factor out ⅓√u
∫ x³ / √(1  x²) dx = ⅓(√u) [ u − 3] + C
Rearrange:
∫ x³ / √(1  x²) dx = ⅓) [ u − 3] (√u) + C
Back subbing u = 1  x²
∫ x³ / √(1  x²) dx = ⅓ [1 x² 3] [√(1x²)] + C
∫ x³ / √(1  x²) dx = ⅓ [ x² 2] √(1x²) + C
Factor out 1:
∫ x³ / √(1  x²) dx = −⅓ [ x² + 2] √(1x²) + C
Now we are ready to apply this to original integral:
∫ (x² ) arcsin x dx = ⅓x³ arcsin x − ⅓∫x³/√(1x²) dx
∫ (x² ) arcsin x dx = ⅓x³ arcsin x − (⅓)( −⅓) [ x² + 2] √(1x²) + C
∫ (x² ) arcsin x dx = ⅓x³ arcsin x + (¹⁄₉) [ x² + 2] √(1x²) + C
If you like, to make this a little more simple, factor out ¹⁄₉
∫ (x² ) arcsin x dx = ¹⁄₉ { 3 x³ arcsin x + ( x² + 2) √(1x²)} + C
after getting ⌠2^u du , can't we use the power rule?
⌠2^u du = (2^(u+1))/(u+1) = 2^(ln(x)+1)/ln(x)+1
⌠2^u du = (2^(u+1))/(u+1) = 2^(ln(x)+1)/ln(x)+1
```
∫(a^x)dx
u = a^x
(d/dx(a^x))dx = du
(d/dx(e^(ln(a^x))))dx = du
(d/dx(e^(x•ln(a))))dx = du
(e^(x•ln(a))•d/dx(x•ln(a)))dx = du
(e^(x•ln(a))•ln(a)•d/dx(x))dx = du
(e^(x•ln(a))•ln(a)•dx/dx)dx = du
(e^(x•ln(a))•ln(a))dx = du
(e^(ln(a^x))•ln(a))dx = du
(a^x•ln(a))dx = du
dx = 1/(a^x•ln(a))du
∫(u/(a^x•ln(a)))du
∫(u/(u•ln(a)))du
1/ln(a)•∫(u/(u))du
1/ln(a)•∫(u^0)du
1/ln(a)•1/(0 + 1)•u^(1 + 0)
1/ln(a)•1/(1)•u^(1)
u/ln(a)
u = a^x
[a^x/ln(a)]
```
∫(a^x)dx
u = a^x
(d/dx(a^x))dx = du
(d/dx(e^(ln(a^x))))dx = du
(d/dx(e^(x•ln(a))))dx = du
(e^(x•ln(a))•d/dx(x•ln(a)))dx = du
(e^(x•ln(a))•ln(a)•d/dx(x))dx = du
(e^(x•ln(a))•ln(a)•dx/dx)dx = du
(e^(x•ln(a))•ln(a))dx = du
(e^(ln(a^x))•ln(a))dx = du
(a^x•ln(a))dx = du
dx = 1/(a^x•ln(a))du
∫(u/(a^x•ln(a)))du
∫(u/(u•ln(a)))du
1/ln(a)•∫(u/(u))du
1/ln(a)•∫(u^0)du
1/ln(a)•1/(0 + 1)•u^(1 + 0)
1/ln(a)•1/(1)•u^(1)
u/ln(a)
u = a^x
[a^x/ln(a)]
```
what exactly are the rules for differentiating trig functions because my lecture solve them differently from khan, for instance how could the rules be applied when we differentiate y=(sinx) to the power of another trig function
There isn't a special set of rules for trig function differentiation. Unlike the math you had in algebra, there is not one set way to do calculus problems. Many problems have multiple ways the can be solved. As a consequence, you can have extremely differentlooking answers (depending on how you chose to solve the problem)  and this is especially true of trig problems. So, just because two answers to the same problem look completely different doesn't necessarily mean that either of them is wrong  they could be just very different ways of expressing the same function.
Integration, is even more complicated as there are integrals which have no known solution. For example, `∫ x^x dx` has no known solution amongst standard mathematical functions. Likewise, there is (as far as I am aware) no known solution (amongst standard mathematical functions) for the integral of a trig function raised to the power of a trig function.
But, if you'd like me to show you how to solve the derivative of a trig function raised to the power of another trig function, here is how to do it:
Let f and g both be trig functions of the variable x.
y = f ^ g
log y = log (f^g)
log y = g log f
Differentiate:
(1/y) y' = (g/f) f' + (log f) g' ← *using the product rule*
y' = y { (g/f) f' + (log f) g' ]
But we know y from the original function, `y= f^g`
y' =[f^g]*{ (g/f) f' + (log f) g' ]
Note: this method works for any two functions of x, not just trig functions.
Thus, the derivative of (sin x)^(cos x) would be:
y' = [(sin x)^(cos x)]*{ ((cos x)/(sin x))(cos x) + (log (sin x))(− sin x) ]
Which simplifies to:
y' = [(sin x)^(cos x)]* [cot (x) cos (x) − sin (x) log (sin x)]
Integration, is even more complicated as there are integrals which have no known solution. For example, `∫ x^x dx` has no known solution amongst standard mathematical functions. Likewise, there is (as far as I am aware) no known solution (amongst standard mathematical functions) for the integral of a trig function raised to the power of a trig function.
But, if you'd like me to show you how to solve the derivative of a trig function raised to the power of another trig function, here is how to do it:
Let f and g both be trig functions of the variable x.
y = f ^ g
log y = log (f^g)
log y = g log f
Differentiate:
(1/y) y' = (g/f) f' + (log f) g' ← *using the product rule*
y' = y { (g/f) f' + (log f) g' ]
But we know y from the original function, `y= f^g`
y' =[f^g]*{ (g/f) f' + (log f) g' ]
Note: this method works for any two functions of x, not just trig functions.
Thus, the derivative of (sin x)^(cos x) would be:
y' = [(sin x)^(cos x)]*{ ((cos x)/(sin x))(cos x) + (log (sin x))(− sin x) ]
Which simplifies to:
y' = [(sin x)^(cos x)]* [cot (x) cos (x) − sin (x) log (sin x)]
How to integrate x^2 (ln x)^2
This requires integrating by parts two separate times. Here's how to do it.
∫ x² ln²(x) dx
Integrate by parts: ∫ fdg = fg ∫ gdf, where
f = ln²(x), dg = x²dx,
df = [2 ln(x)]/x dx, g = ⅓x³
= ⅓ ∫ 2 x² ln(x) dx+⅓ x³ ln²(x)
Factor out 2 from integrand:
= ⅔ ∫ x² ln(x) dx+⅓ x³ ln²(x)
Integrate ∫x² ln(x), by parts, ∫ f dg = f g ∫ g df, where
f = ln(x), dg = x² dx,
df = 1/x dx, g = ⅓x³
= (2 ∫ x² dx)/9²⁄₉ x³ ln(x)+⅓ x³ ln²(x)
=(²⁄₉ ∫ x² dx) ²⁄₉ x³ ln(x)+⅓ x³ ln²(x)
The integral of x² is ⅓x³
= (2 x³)[¹⁄₂₇] ²⁄₉ x³ ln(x)+⅓ x³ ln²(x) + C
You could probably leave it there or you can simplify by factoring out ¹⁄₂₇ x³ :
= [¹⁄₂₇] x³ { 2  6ln(x) + 9ln²(x)} + C
∫ x² ln²(x) dx
Integrate by parts: ∫ fdg = fg ∫ gdf, where
f = ln²(x), dg = x²dx,
df = [2 ln(x)]/x dx, g = ⅓x³
= ⅓ ∫ 2 x² ln(x) dx+⅓ x³ ln²(x)
Factor out 2 from integrand:
= ⅔ ∫ x² ln(x) dx+⅓ x³ ln²(x)
Integrate ∫x² ln(x), by parts, ∫ f dg = f g ∫ g df, where
f = ln(x), dg = x² dx,
df = 1/x dx, g = ⅓x³
= (2 ∫ x² dx)/9²⁄₉ x³ ln(x)+⅓ x³ ln²(x)
=(²⁄₉ ∫ x² dx) ²⁄₉ x³ ln(x)+⅓ x³ ln²(x)
The integral of x² is ⅓x³
= (2 x³)[¹⁄₂₇] ²⁄₉ x³ ln(x)+⅓ x³ ln²(x) + C
You could probably leave it there or you can simplify by factoring out ¹⁄₂₇ x³ :
= [¹⁄₂₇] x³ { 2  6ln(x) + 9ln²(x)} + C
Couldn't the answer also be ((2^lnx)/(ln2)) + C ?
Yes, that's just a different way of writing the solution Sal demonstrates in this video.
at 9:47; is ever number connected to 'ln' and raised to the power 'e' is equal to itself?
I think he does this around 3:17
And yes
a = e^(ln a)
for a > 0
And yes
a = e^(ln a)
for a > 0
The reason, intuitively, for that, is pretty simple.
`log_10 100 = 2` why? because `10^2 = 100`
So, that means `10^(log_10 100) = 10^2` too.
So based on that I can say, `100 = 10^(log_10 100)`.
In variable form, `a = constant, as 100,` and `In = log_e` where e is, Euler constant, 2.17... I could change `log_10` to `log_e` and `10 to e`.
`a= e^In a`, not so difficult.
I hope it helps you.
`log_10 100 = 2` why? because `10^2 = 100`
So, that means `10^(log_10 100) = 10^2` too.
So based on that I can say, `100 = 10^(log_10 100)`.
In variable form, `a = constant, as 100,` and `In = log_e` where e is, Euler constant, 2.17... I could change `log_10` to `log_e` and `10 to e`.
`a= e^In a`, not so difficult.
I hope it helps you.
Is it also correct to rewrite 2^(lnx)=e^(ln(2)*ln(x))=e^*ln(x+2)=x+2?
This gives you the integral to be int[((2+x)/x) dx], which then transcribes to int[(2/x+1)dx], so the solution would be 2ln(x) + x + c.
Is this a proper solution or did I make a mistake?
This gives you the integral to be int[((2+x)/x) dx], which then transcribes to int[(2/x+1)dx], so the solution would be 2ln(x) + x + c.
Is this a proper solution or did I make a mistake?
You made mistakes. Most notably
ln (2)*ln(x) is NOT ln(x+2)
You have that property backward. The correct form is:
ln (ab) = ln (a) + ln (b)
However, it is true that:
ln (a) * ln (b) = ln [a^(ln b)] = ln [b^(ln a)]
And, thus, it is true that:
2^(ln x) = x^(ln 2)
ln (2)*ln(x) is NOT ln(x+2)
You have that property backward. The correct form is:
ln (ab) = ln (a) + ln (b)
However, it is true that:
ln (a) * ln (b) = ln [a^(ln b)] = ln [b^(ln a)]
And, thus, it is true that:
2^(ln x) = x^(ln 2)
at 3:19 can't we directly put the formula int{ a^x}=a^x/lna
Of course we could. I think the reason he did not do that, and rather derived it, was to make sure that everyone can follow along. I am convinced that the sudden appearance of the formula `∫ aᵡ dx = (1 / log a) · aᵡ + C` would confuse _somebody_.
I don't get what he did from 3:36 onwards. Could someone please explain?
When I use int[] I mean to say the indefinite integral of whatever is in the [] brackets.
At 3:36 we have int[2^u du] which looks relatively simplified but we aren't too sure as to what that is, but we are familiar with integrals involving the powers of e.
So a nice way to rewrite 2 is to raise e to the ln(2) power. Because ln(2) gives us the value "what we have to raise e by to get 2", so if we raise e to that power, we're logically going to get 2. So e^ln(2) = 2. Therefore we can rewrite int[2^u du] as:
int[(e^ln(2))^ u du]. By exponent properties, it is equivalent to: [int e^(ln(2) * u) du]
We know that int[e^au du] = 1/a * e^au (There is proof of this in one of the top comments). So, the current integral that we have can be rewritten as:
1/ln(2) * (e^(ln(2) * u)). As was established earlier, e^ln(2) = 2. so e^ln(2)*u = (e^ln(2))^u = 2^u.
So in the end we have 1/ln(2) * 2^u. Now we plug back what u is and we get
1/ln(2) * 2^(ln(x))
At 3:36 we have int[2^u du] which looks relatively simplified but we aren't too sure as to what that is, but we are familiar with integrals involving the powers of e.
So a nice way to rewrite 2 is to raise e to the ln(2) power. Because ln(2) gives us the value "what we have to raise e by to get 2", so if we raise e to that power, we're logically going to get 2. So e^ln(2) = 2. Therefore we can rewrite int[2^u du] as:
int[(e^ln(2))^ u du]. By exponent properties, it is equivalent to: [int e^(ln(2) * u) du]
We know that int[e^au du] = 1/a * e^au (There is proof of this in one of the top comments). So, the current integral that we have can be rewritten as:
1/ln(2) * (e^(ln(2) * u)). As was established earlier, e^ln(2) = 2. so e^ln(2)*u = (e^ln(2))^u = 2^u.
So in the end we have 1/ln(2) * 2^u. Now we plug back what u is and we get
1/ln(2) * 2^(ln(x))
has this type of question any general form ? an easy and convenient way for solving it down in multiple choice questions ?
instead of rewriting it in terms of e, could we not use the integration rule that the integral of a^x=a^x/lna
At 6:36 Sal explains that u ln2 = ln2^u. Would it not be easier to make the next step
(1/ln2)e^u ln2 = e^u (ln2/ln2) =e^u ?
(1/ln2)e^u ln2 = e^u (ln2/ln2) =e^u ?
Why is the problem expressed as (2^ln x)/x, when this is equivalent to (x^ln 2)/x = x^(ln 2  1), which is simpler?
My change from (2^ln x)/x to x^(ln(2)  1) is mostly just the power rules:
2^(ln x) = (e^ln 2)^(ln x) = e^((ln 2) * (ln x)) = (e^ln x)^(ln 2) = x^(ln 2)
(x^(ln 2))/x = (x^(ln 2)) * x^(1) = x^(ln(2)  1)
WolframAlpha verifies that this is true: http://www.wolframalpha.com/input/?i=is+%282%5E%28ln+x%29%29%2Fx+%3D+x%5E%28ln%282%29++1%29
2^(ln x) = (e^ln 2)^(ln x) = e^((ln 2) * (ln x)) = (e^ln x)^(ln 2) = x^(ln 2)
(x^(ln 2))/x = (x^(ln 2)) * x^(1) = x^(ln(2)  1)
WolframAlpha verifies that this is true: http://www.wolframalpha.com/input/?i=is+%282%5E%28ln+x%29%29%2Fx+%3D+x%5E%28ln%282%29++1%29
What exact rule are you using to say that (x^ln 2)/x = x^(ln 2  1)? If this was set up in any form of addition or multiplication where the log rules apply, I could see some of those things swapping around. However, because this is something to a log over something that does not involve a log, I do not see how you are saying they are equal.
Sal, when you get the integral of 2^u dx, can't you just use the reverse power rule and get (2^(u+1))/(u+1)?
No, the reverse power rule is for when the variable is the base, not the exponent.
isn't that 2^u is just 2^u/ln2 by simple way
when Sal gets to 1/(ln 2) * (e^(u ln 2)) + C, can't he just substitute lnx for u, getting
1/(ln 2) * (e^(ln x * ln 2 )) + C, or 1/(ln 2) * ((e^(ln x))^(ln2) + C. Then won't he have 1/ln2 * x^(ln 2) + C?
1/(ln 2) * (e^(ln x * ln 2 )) + C, or 1/(ln 2) * ((e^(ln x))^(ln2) + C. Then won't he have 1/ln2 * x^(ln 2) + C?
when he has `S2^udu`, why can't he just use the reverse product rule??
Can someone please explain me the step Sal takes at 6:19, where he takes the integral? I don't really understand how he got there.
integration of e to the power root x
This problem takes a somewhat strange substitution.
Let u = sqrt(x). Before substituting, notice that u^2 = x.
Thus, 2u du = dx.
Now when you substitute, you get:
integral(e^sqrt(x)) dx = integral(e^u*2u du) = 2integral(u*e^u du).
This integral still requires integration by parts. Hopefully you can proceed from here. If you don't know integration by parts yet, back away from this problem!
Let u = sqrt(x). Before substituting, notice that u^2 = x.
Thus, 2u du = dx.
Now when you substitute, you get:
integral(e^sqrt(x)) dx = integral(e^u*2u du) = 2integral(u*e^u du).
This integral still requires integration by parts. Hopefully you can proceed from here. If you don't know integration by parts yet, back away from this problem!
Can you simplify x^(y^z)?
```e^(ln(x^(y^z)))
e^(y^z•ln(x))```
e^(y^z•ln(x))```
At 3:20 how do we know that we have to rewrite 2^u? Why do you see that it should be in the format of Intergral(e^2)dx?
Are there any methods to solving this particular question?
at 1900, may i please get someone to explain this problem over video chat?
at 4:45 is Sal writing (e^ln2)^u or (e)^ln2*u? I mean e^ln2 equals 2 and 2 is raised to the power u...so why should it be the latter and not the former?
if you remember your logarithmic rules, you will know that log(x^y) = y*log(x)
In the video when you find the integral of (2^u)du. Couldn't you have just said the antiderivative is (2^u) / ln(2) + c without converting to e. I think there is a pretty simple formula that goes like this: The anti derivative of a^x is (a^x) / ln(a). a being any constant like 2 or 3.
The method Sal uses is such a complex way of doing it. Just take u = 2^(lnx). Then dx/x = 1/(uin2), so the integral simplifies to du/ln2 which, once integrated, is equal to u/ln 2 = (2^lnx)/ln2. It's a far easier way of doing it and is simpler because it doesn't have all the e's in it. Using this substitution and method will save you about 5 minutes and a lot of brain power.
03:05 why didn't Sal use rule a^x dx= a^x / ln a +C if a > 0 and a != 0?
it seems to me, he wanted to prove that rule, and he got it right as usual :)
at about 3:30, couldn't you also, or instead, use the fact that since the derivative of 2^u = (ln2)(2^u), the integral of (2^u)(ln2)(du) = 2^u and so the integral of 2^udu = ((2^u)/ln2) + C. That seems a faster way of getting the final answer.
i don't understand how only 1/ln2 came down and nothing else at 6.11 in the video..
At 5:00, why does the integral of e^(au) become 1/a * e^(au)? Where did the 1/a come from? Thanks!
The chain rule, I presume would be the answer, since the expression uses a function of a function ( the thing you wrote up). Hope that helps :)
can you make a video on some standard integral with some examples
This is the playlist for definite and indefinite integrals
https://www.khanacademy.org/math/calculus/integralcalculus
https://www.khanacademy.org/math/calculus/integralcalculus
Is there a way to check your answer after you find the answer from the USubs?
You can check your answer to any kind of integration problem by differentiating your answer. The derivative will equal the original integrand if you did everything correctly. This particular result is somewhat tricky to differentiate, but you can do it if you remember that 1/ln2 is just a constant and 2^lnx can be rewritten as e^(ln2lnx).
Color of the constant is constantly changing :)
what would be the integral of a^x^2 dx??
u could have used the standard integral of a^x=a^x/lna
This sint specific to this video but it does have to do with integration, is it alowable to multiply in a constant into a term your integrating for example if you have ,
7 S (cos(33x))/2 + (cos(66x)*cos(33x))/2
could you multiply the 7 and the bit your trying to integrate by 2 to get
14 S cos(33x) + (cos(33x)*cos(66x))
which doesnt seem equivalent to
7/2 S cos(33x) + (cos(33x) + cos(66x))
so my hunch is no, but why is that the case, since you can pull out a constant of the integral why can't you multiply in a constant?
7 S (cos(33x))/2 + (cos(66x)*cos(33x))/2
could you multiply the 7 and the bit your trying to integrate by 2 to get
14 S cos(33x) + (cos(33x)*cos(66x))
which doesnt seem equivalent to
7/2 S cos(33x) + (cos(33x) + cos(66x))
so my hunch is no, but why is that the case, since you can pull out a constant of the integral why can't you multiply in a constant?
Well,you must understand that when you multiply with 2 the expression becomes 14/4cos(33x)+(cos33x * cos66x).Actually when you "pull out a constant "its already there,which doesn't effect the equality (of the expression) but when you multiply with a constant you must also divide the same constant in order to preserve equality.
Hope that helps
Hope that helps
People have already said something about this, but once you get to e^(u*ln2) you can write this in a few different ways.
Since u=lnx, this is equivalent to
(e^ln x )^ln 2 = x^ln 2
and
(e^ln 2)^ln x = 2 ^ln x.
Since u=lnx, this is equivalent to
(e^ln x )^ln 2 = x^ln 2
and
(e^ln 2)^ln x = 2 ^ln x.
Am I correct in assuming instead of using e^ln2 we could have just multiplied the integral by 1/2 and then multiplied it by e, to make 2 into e, and put an "2*(1/e)" on the outside to distribute back in at the end?
Also in this problem, couldn't you just have found the integral of 2^u which is just 2^u/(ln2)? Or is that incorrect?
Nope that's totally fine if you've memorized that the integral of b^x is (b^x)/ln(b). However it does pay to understand the math behind why that's true. The trick of writing b^x as e^(xlnb) is what derives that formula (and what Sal does in the video for b = 2), but that trick is used all the time in mathematics, not just for this specific family of integral.
What is the antiderivative of something like 2^x? Is it just 2^x/(ln2)?
2^x=e^(ln(2^x))
=> e^(x*ln(2))
Derive => e^(x*ln(2)) * d/dx (x*ln(2))
=> e^(x*ln(2))*ln(2)
=> 2^x*ln(2).
Since ln(2) is a constant, just divide by it again to get back to the original. The antiderivative is 2^x/(ln(2))+C.
=> e^(x*ln(2))
Derive => e^(x*ln(2)) * d/dx (x*ln(2))
=> e^(x*ln(2))*ln(2)
=> 2^x*ln(2).
Since ln(2) is a constant, just divide by it again to get back to the original. The antiderivative is 2^x/(ln(2))+C.
Hi!
I just wanna ask, that why its needed to use the form of the e^ln2? Because already there is a form saying :int a^x dx= (a^x/ln a)+C
I just wanna ask, that why its needed to use the form of the e^ln2? Because already there is a form saying :int a^x dx= (a^x/ln a)+C
can someone please explain why what the logic is in rewriting 2^u to e^ln 2?
Uhhh... I think you missed part. He did not write 2^u as e^ln 2. He wrote it as (e^ln 2)^u, which lets use exponent and log rules to rewite it into a something that has an easy to evaluate item in it: e. Then, once the integral of e is done, he simplifies it back to gets the 2^u/(ln 2).
what does that big line stand for
The yellow one he draws by alnb = lnb^a? That's just showing that the two things are entirely separate, do not mix them.
do you mean the integral sign? It means the counter derivative  what you need to derive in order to get the thing inside the line
hey i got another result for the integral:
(x^(ln 2))/(ln 2) + c
and when i plot the function in wolfram alpha it shows the exact same function as when I plot
sal's result!! Did anyone else got the same result?
(x^(ln 2))/(ln 2) + c
and when i plot the function in wolfram alpha it shows the exact same function as when I plot
sal's result!! Did anyone else got the same result?
can i get someone to explain this in video chat please?
Couldn't you just notice that d/dx 2^ln x= (ln(2)*2^ln(x))/x and correct for the ln(2) to get the antiderivative of (2^ln(x))/x is (2^ln x)/ln(2)?(and to stave off some pointless comments, ln(2)is a constant.)
what is the antiderivetive of lnsec(x)dx?
nevermind. I should watch the video before I ask questions.
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