Integrating Arctan!

**UBCBOY** · February 18th 2013, 10:17 AM

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!

..

**MarkFL** · February 18th 2013, 10:42 AM

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.

**UBCBOY** · February 18th 2013, 11:02 AM

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?

**MarkFL** · February 18th 2013, 11:20 AM

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

**UBCBOY** · February 18th 2013, 12:18 PM

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?

**MarkFL** · February 18th 2013, 12:33 PM

I would leave the second term as:

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

You should find with polynomial division:

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

**ILikeSerena** · February 18th 2013, 12:44 PM

Sorry to interrupt, but I think that:

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

**MarkFL** · February 18th 2013, 12:48 PM

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!

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