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I have been given a few integrals to solve by my math teacher in order to prepare for a competition. In particular, I have some issues with the following integral (which I cannot evaluate by using Symbolab either).

$$ \int \frac{x^2\arctan x}{1+x^2}\ \mathrm{d}x$$

Is there some trick to the question which I have been posed? In any case, could you give me some hints about how to proceed?

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closed as off-topic by Travis, Did, Davide Giraudo, Adam Hughes, Daniel W. Farlow Apr 2 '15 at 18:40

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    $\begingroup$ You can always check if the antiderivative of a function exists in elementary functions by putting it on Wolfram|Alpha. $\endgroup$ – Prasun Biswas Apr 2 '15 at 16:00
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    $\begingroup$ have a look at $u = \arctan x$ as a sub? $\endgroup$ – Chinny84 Apr 2 '15 at 16:01
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    $\begingroup$ Looks pretty doable to me. Note that $x^2\arctan x=(x^2+1)\arctan x-\arctan x$. $\endgroup$ – André Nicolas Apr 2 '15 at 16:01
  • $\begingroup$ @Chinny84 Yes i have and it leaves me with $$1/(1+x^2)$$ which later leaves me with an extra x... $\endgroup$ – LiGNUx Apr 2 '15 at 16:04
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$$\int \frac{x^2 arctanx }{1+x^2}dx=\\\int \frac{(x^2+1-1) arctanx }{1+x^2}dx=\\ \int (1-\frac{1}{1+x^2})arctan x dx=\\\int arctanx dx - \int \frac{1}{1+x^2}arctanx dx$$ for the first one use integration by part , and for second use $u=arctanx\\du=\frac{1}{1+x^2}dx$ $$\int arctanx dx =x arctanx -\int x \frac{1}{1+x^2} dx=\\x arctanx -\frac{1}{2}ln(1+x^2)\\$$ $$\int\frac{arctan x}{1+x^2}dx=\int u du=\frac{u^2}{2}=\frac{arctan^2 x}{2} $$

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Not wanting to answer questions in teh comments $$ u = \arctan x \implies du = \frac{1}{1+x^2}dx $$ thus $$ \int \frac{x^2\arctan x}{1+x^2}dx = \int x^2 udu = \int u\tan^2u du $$ thus you can use integration by parts to solve the rest?

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  • $\begingroup$ Yes, you'll need to use the identity $\sec^2(u)-\tan^2(u)=1$ to ease out the computations of the integral and then integrate $f(u)=\tan(u)$ using IBP again to complete. $\endgroup$ – Prasun Biswas Apr 2 '15 at 16:09
  • $\begingroup$ good to see you man. $\endgroup$ – abel Apr 2 '15 at 16:22
  • $\begingroup$ Ah @abel always a pleasure. Keeping well I hope. $\endgroup$ – Chinny84 Apr 2 '15 at 16:25
  • $\begingroup$ i am good. spring break! how did your interviews go? $\endgroup$ – abel Apr 2 '15 at 16:26
  • $\begingroup$ I hope your not "drinking and solving" ;). As for interviews not too bad, just waiting for feedback mostly. $\endgroup$ – Chinny84 Apr 2 '15 at 16:29
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You say (in the tags) that it's a definite integral, but you don't provide the integration limits.

If by chance you're integrating from minus infinity to plus infinity (or $-p$ to $+p$ for any value of p), then you can use the fact that arctan is an odd function: that is, $arctan(-x) = - arctan(x)$.

As $x^2$ is an even function, the whole integrand is odd.

Then the integral over negative $x$ cancels out the integral over positive $x$, and you wind up with zero.

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  • $\begingroup$ I now notice the "definite-integrals" tag has been removed, so I presume this answer is actually not of any use. $\endgroup$ – IanF1 Apr 2 '15 at 16:14

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