Integration techniques

# u-substitution

## (2^(ln x))/x antiderivative example

Finding  ∫(2^ln x)/x dx
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## (2^(ln x))/x antiderivative example

Discussion and questions for this video
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)]
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.
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.
Why is this video in the Multivariable Calculus playlist?
Probably irrelevant by now, but it's been moved to the U-substitution section because the solution is found using U-substitution.
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.
1 Comment
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
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.
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.
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.
1 Comment
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.
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 :)
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)
[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.
1 Comment
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 u-substitution 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 u-substitution using ln x here because the formula also includes 1/x, the derivative of ln x. We can't do a u-substitution 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/√(1-x²)
Let dv = x²
Thus v = ⅓x³
Thus,
∫ (x² ) arcsin x dx = ⅓ x³ arcsin x − ⅓∫x³/√(1-x²) 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] [√(1-x²)] + C
∫ x³ / √(1 - x²) dx = ⅓ [- x² -2] √(1-x²) + C
Factor out -1:
∫ x³ / √(1 - x²) dx = −⅓ [ x² + 2] √(1-x²) + C
Now we are ready to apply this to original integral:
∫ (x² ) arcsin x dx = ⅓x³ arcsin x − ⅓∫x³/√(1-x²) dx
∫ (x² ) arcsin x dx = ⅓x³ arcsin x − (⅓)( −⅓) [ x² + 2] √(1-x²) + C
∫ (x² ) arcsin x dx = ⅓x³ arcsin x + (¹⁄₉) [ x² + 2] √(1-x²) + 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) √(1-x²)} + 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

∫(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 different-looking 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)]
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)/9-²⁄₉ 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
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?
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))
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 ?
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
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?
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!
Can you simplify x^(y^z)?
e^(ln(x^(y^z)))

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 anti-derivative 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

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?
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
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.
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
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.
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?
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 ln|sec(x)|dx?