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## 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
Homework Helper

[/tex]

## 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

## The Attempt at a Solution

Thank You! I know I can solve it now!