Just, the Integral of (x^3)/(x^3+1) dx

I know you can turn the denominator into (X+1)(X^2-x+1). What do you do from there? thanks =)

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- Sep 24th 2008, 01:34 PM3deltatOne integration by parts problem
Just, the Integral of (x^3)/(x^3+1) dx

I know you can turn the denominator into (X+1)(X^2-x+1). What do you do from there? thanks =) - Sep 24th 2008, 04:55 PMdax918
I doublechecked this but I'm a bit rusty lately so perhaps someone else can verify it,

int dx/(x+1)(x^2-x+1)

You have an irreducible quadratic factor so the equation is

a/(x+1) + (bx+c)/(x^2-x+1) = 1

x=0 -> a+c=1

x=1 -> a+2b+2c=1

x=-1 -> 3a=1

Solving that system, a=1/3, b=-1/3, c=2/3

Now all you have to do is to solve the last part:

int 1/(3(x+1)) + ((-1/3)x + (2/3))/(x^2-x+1)

Hope this helps :) - Sep 24th 2008, 05:10 PMJhevon
**dax918**is right for the part he did, but he left out an important detail. your integral is actually the integral of 1 minus all he said.

Note that $\displaystyle \frac {x^3}{x^3 + 1} = \frac {x^3 + 1 - 1}{x^3 + 1} = \frac {x^3 + 1}{x^3 + 1} - \frac 1{x^3 + 1} = 1 - \frac 1{x^3 + 1}$

as dax918 said, $\displaystyle \frac 1{x^3 + 1} = \frac 13 \bigg( \frac 1{x + 1} - \frac {x - 2}{x^2 - x + 1} \bigg)$

thus, your integral is $\displaystyle \int \Bigg[ 1 - \frac 13 \bigg( \frac 1{x + 1} - \frac {x - 2}{x^2 - x + 1} \bigg) \Bigg]~dx$ - Sep 24th 2008, 05:31 PMdax918
Thank you Jhevon for pointing that out, I forgot the fact that 3deltat already did the first step :)