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When doing differentiation, I know that if $f$ is a function on $x$, then

$$ { d \over dx } f^2 = 2 f {df \over dx} $$

so the opposite in integration is also clear:

$$ \int 2 f { df \over dx } dx = f^2 $$

I also know that

$$ \int x^2 dx = { x^3 \over 3} $$

But I'm not sure as to how I can evaluate:

$$ \int f^2 dx $$

I mean is there any identity for this? That the above is equal to another function of $f$ (such as $f^3 \over 3$ times something)? Is there any method to find this? I googled some but perhaps I wasn't using proper search terms so I didn't get any clear results so I'm asking here. [I hope my question is clear enough :-(]

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Can't be done, in general. For example, it is easy to do $$\int xe^{x^2}\,dx$$ but there is no expression for $$\int x^2e^{2x^2}\,dx$$ in terms of the familiar functions of undergraduate mathematics.

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Hi thanks for replying. I'd like a couple of clarifications. I presume it is easy to evaluate $$ \int x e^(x^2) dx $$ because it can be rearranged to be a product of the chain rule as $$ \int {2xe^{x^2} \over 2} dx $$ whereas the second example cannot. But I am intrigued by your remark on "undergraduate" mathematics. Is there anything in higher level mathematics that can express the integral of the second example? – jamadagni Nov 26 '12 at 6:45
Yes --- and, no. In higher level studies you just define a new function, the "error function", by ${\rm erf}(x)=\int_{-\infty}^xe^{-t^2}\,dt$ (or something like that), and then you can express the second integral in terms of erf. – Gerry Myerson Nov 26 '12 at 10:05

i suppose if you substitute the function as any arbitrary variable X u can solve this integration. for example while integrating (1+x)^2 you substitute 1+x=t then differentiate both with respect to t. this gives you dx=dt. here, no matter what your function, make dx the subject of the equation so that you can replace dx in your original integral. after that you integrate the function as integration of t^2 which is t^3/3 and then resubstitute t=1+x in the answer to get your answer.This is of course a very simple example but the process is the same.

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sorry i don't know how to format it so that i could show you a proper solved example with the integral sign and everything – chinar May 10 '14 at 21:32
This is a good idea that will probably work in some cases. I think people are downvoting because it's just a trick that might work, not a general solution. – 6005 May 11 '14 at 17:40
it works in most cases where the function isn't really complicated. i really think the person with the query should at least try this.... – chinar May 25 '14 at 14:29

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