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 :-(]


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.

  • $\begingroup$ 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? $\endgroup$ – jamadagni Nov 26 '12 at 6:45
  • 6
    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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 $\endgroup$ – chinar May 10 '14 at 21:32
  • $\begingroup$ 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. $\endgroup$ – 6005 May 11 '14 at 17:40
  • $\begingroup$ 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.... $\endgroup$ – chinar May 25 '14 at 14:29
  • $\begingroup$ I think this solution only works if the inner function is linear $\endgroup$ – Álvaro Paz Jun 30 '20 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.