How do I solve an integral using partial fractions?

[Davidk1]

Please help me integrate homework :(

[/tex]

Homework Statement

Homework Equations

\int dx/(1-x^2)^3

The Attempt at a Solution

I believe I have to use partial fractions to solve this integral. I started out by expanding the denominator

1/-(1 - 3x^2 + 3x^4 - x^6)

I pulled out a negative then separated by polynomial

x^4 (x^2 + 3) - 3(x^2 + 3) - 8

so:
-(x^4 - 3)(x^2 + 3) - 8

I'm lost..I need help

 

[HallsofIvy]

[/tex]

Homework Statement

Homework Equations

\int dx/(1-x^2)^3

The Attempt at a Solution

I believe I have to use partial fractions to solve this integral. I started out by expanding the denominator

1/-(1 - 3x^2 + 3x^4 - x^6)

Why in the world would you expand it? You want to factor it. It is already partly factored, don't throw that away! 1- x²= (1- x)(1+ x) so
(1-x²)³= (1- x)³(1+ x)³

$$\frac{1}{(1- x^2)^3}= \frac{A}{1-x}+ \frac{B}{(1-x)^2}+ \frac{C}{(1- x)^3}+ \frac{D}{1+ x}+ \frac{E}{(1+x)^2}+ \frac{F}{(1+x)^3}$$

I pulled out a negative then separated by polynomial

x^4 (x^2 + 3) - 3(x^2 + 3) - 8

so:
-(x^4 - 3)(x^2 + 3) - 8

I'm lost..I need help

 

[Davidk1]

Thank You! I know I can solve it now!