Problem resolving an Integral - Partial Fractions

[SclayP]

1. So, i have the next integrand...
2. $\int \frac{1}{(x-1)^2(x+1)^2}\,dx$
3. I proceeded by resolving it by partial fraction and i came up with the next...

$\int \frac{1}{((x-1)^2)((x+1)^2)}\,dx = \int \frac{A}{(x-1)} + \frac{B}{(x-1)^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2}\,dx$

The thing is that after doing all the calculus i came up with this...

$A + C = 0$
$A + B - C + D = 0$
$-A + 2B - C -2D = 0$
$A + B + C + D = 1$

After this i don't know how to preceed i mean i don't know how to resolve the equation whit 4 variables...

Thanks, and very sorry for my english...

 

[SammyS]

1. So, i have the next integrand...

2. $\int \frac{1}{(x-1)^2(x+1)^2}\,dx$

3. I proceeded by resolving it by partial fraction and i came up with the next...

$\int \frac{1}{((x-1)^2)((x+1)^2)}\,dx = \int \frac{A}{(x-1)} + \frac{B}{(x-1)^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2}\,dx$

The thing is that after doing all the calculus i came up with this...

$A + C = 0$
$A + B - C + D = 0$
$-A + 2B - C -2D = 0$
$A + B + C + D = 1$

After this i don't know how to proceed i mean i don't know how to resolve the equation whit 4 variables...

Thanks, and very sorry for my english...

Use $\ \ A + C = 0\ \$ with $\ \ -A + 2B - C -2D = 0\ \$

to get $\ \ 2B -2D = 0\ .$

Similarly, use $\ \ A + C = 0\ \$ with $\ \ A + B + C + D = 1$

to get $\ \ B + D = 1\ .$

Use those two equations to solve for B & D .

Put the results for B & D into the first two equations to get A & C .

 

[SammyS]

As an alternative to the above method, consider the following. (This method was also discussed in the thread: Partial Fractions)

I assume that you got your 4 equations for A, B, C, and D by equating coefficients for powers of x in the following equation.

$1=A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2)(x+1)+D(x-1)^2\ .$

Your can quickly solve for B, by letting x = 1 .

Similarly, you can solve for D, by letting x = -1 .

After that it's not so difficult to find A & C .