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## Homework Statement

use the method of partial fractions on [tex]\int[/tex] [tex]\frac{36}{(x-2)(x-1)^2(x+1)^2}[/tex] dx

**2. Homework Equations**

**3. The Attempt at a Solution**

[tex]\frac{A}{(x-2)}[/tex] + [tex]\frac{B}{(x-1)}[/tex] + [tex]\frac{C}{(x-1)^2}[/tex] + [tex]\frac{D}{(x+1)}[/tex] + [tex]\frac{E}{(x+1)^2}[/tex]

I took the 36 out of the integral. Then

A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1

In the first fraction (A) (all terms cancel)

I tried to plug in x=

1, -1, 0, 2, -2, into (A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1)

so...

when x=1 I ended up with B(-2) +C(-4)=1 (I know this is probably simple algebra or I'm definitely screwing up somewhere

Any help is appreciated[tex]\frac{A}{(x-2)}[/tex] + [tex]\frac{B}{(x-1)}[/tex] + [tex]\frac{C}{(x-1)^2}[/tex] + [tex]\frac{D}{(x+1)}[/tex] + [tex]\frac{E}{(x+1)^2}[/tex]

I took the 36 out of the integral. Then

A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1

In the first fraction (A) (all terms cancel)

I tried to plug in x=

1, -1, 0, 2, -2, into (A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1)

so...

when x=1 I ended up with B(-2) +C(-4)=1 (I know this is probably simple algebra or I'm definitely screwing up somewhere

Any help is appreciated