Two integrals giving me a hard time

**rioneye** · October 16th 2009, 06:45 AM

The first one is:

$\int x^5 ln(1+x^3) dx$

The second is:

$\int_0^1 \frac {1}{(16+x^2)^2}dx$

For both problems I used parts but the problems got quite ugly so I stopped. I was wondering if there are any rules or methods I am missing that would make these problems easier

**Scott H** · October 16th 2009, 07:13 AM

The first can be solved integrating by parts and then performing a substitution.

The second contains an expression of the form $a^2+x^2$ . Try substituting $x=a\tan u$ .

**JG89** · October 16th 2009, 07:16 AM

$\int x^5 log(1 + x^3) dx = \int x^2 log(1 + x^3) x^3 dx$ so use the substitution $u = x^3$ .

I haven't tried to work out the second one yet, but when I look at it, the substitution $u = 4tan(x)$ jumps out at me.

**rioneye** · October 16th 2009, 08:10 AM

I'm still having trouble on both. On the first integral I've substituted $u=x^3$ then I did parts and got stuck. On the second one I did trig sub but I get stuck when you plug the original variables back in (what does $arctan \frac {1}{4}$ equal?). Thanks for both your help so far, but I still need a little bit more help.

**rioneye** · October 16th 2009, 01:24 PM

So I think I solved the first one but the second one is still giving me some trouble when it comes to substituting the bounds back in.

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