A way to solve but surely not the best,
1. let x=1/t then ¡Ò(t^3/(t^6+1))dt=-¡Ò(xdx/(x^6+1))
2. let u=x^2 then the integral changes to -0.5*¡Òdu/(u^3+1)
knowing that u^3+1=(u+1)(u^2-u+1) u can solve the integral.

But the process is kinda long and the result looks weird

[edited by - remi on January 24, 2004 7:38:53 AM]

quote:Original post by sadwanmage
Ok, I think I was able to it. It ain''t pretty. My answer differs significantly from the Integrators, but I think either I have made simple mistake, or I am missing some clever re-arrangement that makes them the same. So I am fairly sure the following method is along the right lines:

I used the substitution x=t^2. From there its partial fractions, and then a reach to the formula booklet for the last part. A nice integral, it covers practically everything I have learnt on integration. It does involve lns and arctans.

I''m going to give your method a whirl. If I substitute x = t^2 I should get a sum of perfect cubes on the bottom to work with and then split that apart with partial fractions..sounds like a plan. I really hope the teacher has no intention of putting a question like that on the test. In fact, I have covered partial fractions in precalculus a while back, but I think there is a special section devoted to using partial fractions later on in this calculus II book, so I''d like to think she''s not dumping something on us that should be covered in a later section. Nervertheless, I''m going to attempt to integrate according to your method sadwanmage or remi. Thank you all very much for taking the time to help me.