1. ## Integrating Arctan!

I can't seem to find a substitution for the inverse of tan anywhere, and online it looks like a mess. Can someone tell me the step by step process for integrating the following, please.

x^2arctan(6x)dx

Is this with integration by parts?

Thank you!

..

2. ## Re: Integrating Arctan!

Yes, I would try integration by parts, where:

$u=\tan^{-1}(6x)\,\therefore\,du=\frac{1}{(6x)^2+1}\,dx$

$dv=x^2\,dx\,\therefore\,v=\frac{1}{3}x^3$

Then use long division on the integrand of the integral that results.

3. ## Re: Integrating Arctan!

Originally Posted by MarkFL2
Yes, I would try integration by parts, where:

$u=\tan^{-1}(6x)\,\therefore\,du=\frac{1}{(6x)^2+1}\,dx$

$dv=x^2\,dx\,\therefore\,v=\frac{1}{3}x^3$

Then use long division on the integrand of the integral that results.
Thanks for your reply, just a question by the way, does it matter which term I choose as u or v when I'm integrating by parts?

4. ## Re: Integrating Arctan!

Yes, I use the LIATE rule to decide which will be u.

5. ## Re: Integrating Arctan!

Originally Posted by MarkFL2
Yes, I use the LIATE rule to decide which will be u.
After doing so, I'm left with:

x^3arctan(6x)/3 - ∫x^3/(108x^2 + 3)

(sorry for the messy notation)

How do I go about integrating the second term?

6. ## Re: Integrating Arctan!

I would leave the second term as:

$-\frac{1}{3}\int\frac{x^3}{36x^2+1}\,dx$

You should find with polynomial division:

$\frac{x^3}{36x^2+1}=\frac{1}{36}\left(x-\frac{x}{36x^2+1} \right)$

7. ## Re: Integrating Arctan!

Sorry to interrupt, but I think that:

$d(\arctan(6x)) = \frac {6 dx}{(6x)^2 + 1}$

8. ## Re: Integrating Arctan!

Originally Posted by ILikeSerena
Sorry to interrupt, but I think that:

$d(\arctan(6x)) = \frac {6 dx}{(6x)^2 + 1}$
Thanks, you are right, I got careless with the chain rule!