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Integration techniques

### u-substitution

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## (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.
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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)]
Why is this video in the Multivariable Calculus playlist?
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Probably irrelevant by now, but it's been moved to the U-substitution section because the solution is found using U-substitution.
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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.
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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.
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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
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@ 3:16 I didnt get how' 2 is the same thing as e to the natural log of 2 ' ?can someone please explain?
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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.
how to integrate ln(x)^2
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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
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Yes thats correct. Sal was just explaining the concept without using that formula.
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You could have taken u=2^ln x differentiated and then substituted right?
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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.
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How to integrate x^2 (ln x)^2
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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 ?
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Yes, that's just a different way of writing the solution Sal demonstrates in this video.
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I don't get what he did from 3:36 onwards. Could someone please explain?
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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))
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has this type of question any general form ? an easy and convenient way for solving it down in multiple choice questions ?
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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?
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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
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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)?
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No, the reverse power rule is for when the variable is the base, not the exponent.
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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?
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when he has S2^udu, why can't he just use the reverse product rule??
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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.
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integration of e to the power root x
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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!
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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?
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Are there any methods to solving this particular question?
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at 1900, may i please get someone to explain this problem over video chat?
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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?
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if you remember your logarithmic rules, you will know that log(x^y) = y*log(x)
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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.
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i don't understand how only 1/ln2 came down and nothing else at 6.11 in the video..
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At 5:00, why does the integral of e^(au) become 1/a * e^(au)? Where did the 1/a come from? Thanks!
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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 :)
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can you make a video on some standard integral with some examples
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This is the playlist for definite and indefinite integrals

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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).
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Color of the constant is constantly changing :)
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what would be the integral of a^x^2 dx??
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u could have used the standard integral of a^x=a^x/lna
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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?
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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?
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1 Comment
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.
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What is the antiderivative of something like 2^x? Is it just 2^x/(ln2)?
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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.
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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
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can someone please explain why what the logic is in rewriting 2^u to e^ln 2?
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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).
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what does that big line stand for
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The yellow one he draws by alnb = lnb^a? That's just showing that the two things are entirely separate, do not mix them.
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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?
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can i get someone to explain this in video chat please?
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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.)
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what is the antiderivetive of ln|sec(x)|dx?
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