How do you use partial fractions to find the integral `int (x^3)/(x^2-4)^2dx`?

Perform the decomposition into partial fractions

`x^3/(x^2-4)^2=(x^3)/((x-2)^2(x+2)^2)`

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

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

The denominators are the same, compare the numerators

`x^3=A(x+2)^2+B(x-2)(x+2)^2+C(x-2)^2+D(x+2)(x-2)^2`

Let `x=2`, `=>`, `8=16A`, `=>`, `A=1/2`

Let `x=-2`, `=>`, `-8=16C`, `=>`, `C=-1/2`

Coefficients of `x^3`

`1=B+D`

Coefficients of `x^2`

`0=A+2B+C-2D`

`B-D=0`

`B=B=1/2`

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

Therefore, the integral is

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

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

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

As a Second Method, let us solve the Problem without

applying the Method of Partial Fraction.

Consider the subst. `x^2=y," so that, "2xdx=dy`.

`:. I=intx^3/(x^2-4)^2dx=1/2int(x^2*2x)/(x^2-4)^2dx`,

`=1/2inty/(y-4)^2dy`,

`=1/2int{(y-4)+4}/(y-4)^2dy`,

`=1/2int{(y-4)/(y-4)^2+4/(y-4)^2}dy`,

`=1/2{int1/(y-4)dy+4int1/(y-4)^2dy}`,

`=1/2{ln|(y-4)|+4*(y-4)^(-2+1)/(-2+1)}`.

`=1/2ln|(y-4)|-2/(y-4)`.

Returning to `x^2=y`, we get,

`I=1/2ln|(x^2-4)|-2/(x^2-4)+C`, as Respected Narad T. has

readily derived!

Enjoy Maths.!