# Homework Help: Tough Integral

1. Nov 27, 2008

### the.flea

1. The problem statement, all variables and given/known data
Integral of: 2x^3/(x^3-1) with respect to x.

3. The attempt at a solution
you can divide them to get sum of integrals.
=int(2,x) + 2*int(1/(x^-1),x)
=2x + 2*int(1/(x^-1),x)

Im having trouble proceeding with the last part, i just need help getting started.

Last edited: Nov 27, 2008
2. Nov 27, 2008

### jeffreydk

I think there is an error in your polynomial division if what you are meaning to say is that

$$\frac{2x^3}{x^3-1}=2+2\frac{1}{x^{-1}}=2+2x$$

The way that I would do it is to split up the integral like this..

$$\int\frac{2x^3}{x^3-1}dx=\int x \frac{2x^2}{x^3-1}dx=x\ln{(x^3-1)}-\int \ln{(x^3-1)}dx$$

where you get to the last part of the equality by doing by parts on the split up parts.

By the way you could do the polynomial division but it wouldn't make it much simpler, I think you just get

$$\frac{2x^3}{x^3-1}=\frac{2}{x^3-1}+2$$

Last edited: Nov 27, 2008
3. Nov 27, 2008

### Unco

I realise how ridiculous integral threads can become when everyone wants to chip in with their 'better' method, but I think it's worth mentioning that if you'd rather not integrate $$\ln{(x^3-1)}$$ as Jeff's approach requires, another path is to factorise the original integrand's denominator and use partial fractions.

Show us your work if you get stuck, Mr Flea.

4. Nov 28, 2008

### the.flea

thats the thing i know how to integrate ln(x) where x is any function wih power 1.
ln(x^3-1)
let t = x^3-1 --> 3x^2dx=dt when i sub that in i dont know how to proceed
=ln(t)dt/3x^2 --> maybe another substitution, ie let u=3x^2, du=6xdt,
any help guys?

Last edited: Nov 28, 2008
5. Nov 28, 2008

### Staff: Mentor

Before you start in with partial fractions decomposition, the rational expression should be made into a proper rational expression, one for which the degree of the numerator is less than the degree of the denominator.

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

Now you can factor the denominator in the second term and use partial fractions.