Question

How do you solve this integral?

`intdx/(x^2-1)^2`

Answer 1

`int\ ("d"x)/(x^2-1)^2`
`=1/4(ln(x+1)-ln(x-1)-(2x)/(x^2-1))+C`

Explanation:

`int\ ("d"x)/(x^2-1)^2`
`=int\ ("d"x)/((x+1)^2(x-1)^2)`

Now, let's do the partial fractions. Assume that

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

for some constants `A,B,C,D`.

Then,
`1=A(x+1)(x-1)^2+B(x-1)^2+C(x+1)^2(x-1)+D(x+1)^2`

Expand to get
`1=(A+C)x^3+(B+C+D-A)x^2+(2D-2B-A-C)x+A+B-C+D`.

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

Solving gives `A=B=D=1/4` and `C=-1/4`.

Thus, our original integral is
`int\ (1/(4(x+1))+1/(4(x+1)^2)-1/(4(x-1))+1/(4(x-1)^2))\ "d"x`
`=1/4ln(x+1)-1/(4(x+1))-1/4ln(x-1)-1/(4(x-1))+C`

Simplify:
`=1/4(ln(x+1)-ln(x-1)-(2x)/(x^2-1))+C`