Question

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

Answer 1

The answer is `=-1/2ln(|x^2-1|)+ln(|x|)-1/(2(x^2-1))+C`

Explanation:

Reminder

`x^2-1=(x+1)(x-1)`

Let's perform the decomposition into partial fractions

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

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

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

The denominators are the same, compare the numerators

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

Let `x=0`, `=>`, `1=A`

Let `x=1`, `=>`, `1=4D`, `=>`, `D=1/4`

Let `x=-1`, `=>`, `1=-4B`, `=>`, `B=-1/4`

Coefficients of `x^4`

`0=A+C+E`, `=>`, `C+E=-A=-1`

Coefficients of `x^3`

`0=B+D-C+E`, `=>`, `B+D=C-E=0`

`C=E=-1/2`

Therefore,

`1/(x(x^2-1)^2)=1/x+(-1/4)/(x+1)^2+(-1/2)/(x+1)+(1/4)/(x-1)^2+(-1/2)/(x-1)`

So,

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

`=ln(|x|)+1/(4(x+1))-1/2ln(|x+1|)-1/(4(x-1))-1/2ln(|x-1|)+C`