Question

How do you integrate `2/((x-1)^2(x+1))` using partial fractions?

Answer 1

`int 2/((x-1)^2(x+1)) dx`

`= -1/2 ln abs(x-1)-1/(x-1)+1/2ln abs(x+1) + C`

Explanation:

`2/((x-1)^2(x+1)) = A/(x-1)+B/(x-1)^2+C/(x+1)`

`color(white)(2/((x-1)^2(x+1))) = (A(x-1)(x+1)+B(x+1)+C(x-1)^2)/((x-1)^2(x+1))`

`color(white)(2/((x-1)^2(x+1))) = ((A+C)x^2+(B-2C)x+(-A+B+C))/((x-1)^2(x+1))`

Equating coefficients, we find:

`{ (A+C=0), (B-2C=0), (-A+B+C=2) :}`

Adding all three equations together, we find:

`2B=2`

and hence:

`B=1`

Then from the second equation we find:

`C=1/2`

Then from the first equation we find:

`A=-1/2`

So:

`int 2/((x-1)^2(x+1)) dx`

`= int -1/2(1/(x-1))+1/(x-1)^2+1/2(1/(x+1)) dx`

`= -1/2 ln abs(x-1)-1/(x-1)+1/2ln abs(x+1) + C`