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 .