Question

How do you find the integral `int x^2/((x-1)^2(x+1)) dx` ?

Answer 1

`int x^2/((x-1)^2(x+1)) dx = 3/4 ln abs(x-1)-1/(2(x-1))+1/4 ln abs(x+1) + C`

Explanation:

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

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

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

Equating coefficients:

`{ (A+C=1), (B-2C=0), (-A+B+C=0) :}`

Adding all three equations, we find:

`2B=1`

So:

`B=1/2`

Then from the second equation, we find:

`C=1/4`

Then from the first equation we find:

`A=3/4`

So:

`int x^2/((x-1)^2(x+1)) dx = int (3/4*1/(x-1)+1/2*1/(x-1)^2+1/4*1/(x+1)) dx`

`color(white)(int x^2/((x-1)^2(x+1)) dx) = 3/4 ln abs(x-1)-1/(2(x-1))+1/4 ln abs(x+1) + C`