# Evaluate the following integral: $\int \frac{x^2\arctan x}{1+x^2}\ \mathrm{d}x$ [closed]

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Did, Davide Giraudo, Adam Hughes, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

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

$$\int \frac{x^2 arctanx }{1+x^2}dx=\\\int \frac{(x^2+1-1) arctanx }{1+x^2}dx=\\ \int (1-\frac{1}{1+x^2})arctan x dx=\\\int arctanx dx - \int \frac{1}{1+x^2}arctanx dx$$ for the first one use integration by part , and for second use $u=arctanx\\du=\frac{1}{1+x^2}dx$ $$\int arctanx dx =x arctanx -\int x \frac{1}{1+x^2} dx=\\x arctanx -\frac{1}{2}ln(1+x^2)\\$$ $$\int\frac{arctan x}{1+x^2}dx=\int u du=\frac{u^2}{2}=\frac{arctan^2 x}{2}$$

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?

• 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. – Prasun Biswas Apr 2 '15 at 16:09
• good to see you man. – abel Apr 2 '15 at 16:22
• Ah @abel always a pleasure. Keeping well I hope. – Chinny84 Apr 2 '15 at 16:25
• i am good. spring break! how did your interviews go? – abel Apr 2 '15 at 16:26
• I hope your not "drinking and solving" ;). As for interviews not too bad, just waiting for feedback mostly. – Chinny84 Apr 2 '15 at 16:29

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.

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