How the heck do you integrate this? x dx/ (x^3 - 1) ...... (x^3 means "x cubed") I spent like 3 hours trying to solve this problem but couldn't do it. Thanks in advance.
EDIT.. you did not see the x*dx DIVIDED BY (x^3-1) in any case.. i used an online integrator and that answer that popped up was too complicated.. im not gonna touch this one
From first look, I thought it was partial fractions, but it's not. And obviously substitution doesn't work either. I'm thinking maybe trigonometric substitution, but how do you make it into a^2 + x^2 form? I don't know.
alright nvm ill try, how about trig sub: x^2=sec(x)? then use the identity tan(x)^2=sec(x)^2-1 dunno if that works
Can you elaborate? This is what I got if I substitute like you said: Sqrt (sec x) dx/ ((Sqrt (sec x) * (tan^2(x) - 1)-1) ........ This still doesn't work because how do you take care of the denominator? It's even messier than the original problem.
ahh my bad, its way too early in the morning. I messed up in the beginning. What i wrote does not work at all. Back to the drawing board because i have no clue what to do.
its been a while since ive done this stuff.. im gonna let you try this one if you want, because i have to go x^3-1 factors into (x-1)*(X^2+x+1) there is a trick in integration where x/a= (x-1+1)/a so by applying that (x-1+1)/[(x-1)*(X^2+x+1)] split them up into two different fractions (x-1)/[(x-1)*(X^2+x+1)] + 1/[(x-1)*(X^2+x+1)] in the first term, the (x-1) cancel out you can try this peace.
I thought you needed to use substitution. u = x^3 -1 du = 3x^2 dx Take the 3 out so it becomes x^2 dx / (x^3 -1) or du/u which makes it 3 ln (x^3-1) + C I think that's what the answer is, not entirely sure that it's right.
Actually, I don't think it's right, I had x^2 dx on the top while the question was only x dx. My bad.
On my ti-89 I get: [ln( (x - 1)^2 / abs(x^2 + x + 1) ) / 6] + [ sqrt(3) * invtan(sqrt(3) *(2x + 1)) / 3]
I would guess factoring the x^3-1 into an x^2 form, then using u substitution with the factor being 1/2. I'm too lazy to actually check if that works though...and it's probably a bit too simple considering the answers getting churned out lol.
