How to evaluate this definite integral??
$$ \frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx $$

http://www.wolframalpha.com/input/?i=64%2Fpi^3+*+integrate+from+0+to+infinity+%28ln+x%29^2+%2815-2x%29%2F%28%28x^4%2B1%29%28x^2%2B1%29%29

Question

How to evaluate this definite integral??
$$ \frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx $$

http://www.wolframalpha.com/input/?i=64%2Fpi^3+*+integrate+from+0+to+infinity+%28ln+x%29^2+%2815-2x%29%2F%28%28x^4%2B1%29%28x^2%2B1%29%29

anonymous · Answer

$$ \frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx $$

australopithecus · Answer

Try integration by parts

anonymous · Answer

maybe .. 
but very difficult for me.

australopithecus · Answer

do you not know how to do integration by parts?

anonymous · Answer

no sorry ... could you help me please??

australopithecus · Answer

Nvm integration by parts wouldn't work to my knowledge I think you are going to need someone with a greater understanding of integrals to help you 

@satellite73 @KingGeorge

anonymous · Answer

thank you :)

anonymous · Answer

one method is letting  $\ln x=t$  and representing integrand by series...

@experimentX  come here my friend

experimentx · Answer

huh?

experimentx · Answer

we need something that converges between 0 and infinity http://www.wolframalpha.com/input/?i=expand+%28ln+x%29^2+%2815-2x%29%2F%28%28x^4%2B1%29%28x^2%2B1%29%29+at+x%3D1

experimentx · Answer

need something like |dw:1345610604780:dw| ... let's see what series has closed value of pi^3

anonymous · Answer

the original question was posted here http://math.stackexchange.com/questions/168485/help-find-hard-integrals-that-evaluate-to-59/168509#168509

experimentx · Answer

|dw:1345610982159:dw|