Solving an Integral: 2x^3/(x^3-1)

[the.flea]

Homework Statement

Integral of: 2x^3/(x^3-1) with respect to x.

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.

 

[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

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

 

[Unco]

I realize 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.

 

[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 don't know how to proceed
=ln(t)dt/3x^2 --> maybe another substitution, ie let u=3x^2, du=6xdt,
any help guys?

 

[Mark44]

I realize 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.

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.