Question

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

Answer 1

The partial fraction equation is:

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

Multiply both sides by `x(x+1)^2`

`1 = A(x+1)^2 + Bx(x+1)+Cx`

Eliminate B and C by Letting `x = 0`:

`1 = A(0+1)^2`

`A = 1`

`1 = (x+1)^2 + Bx(x+1)+Cx`

Eliminate B by letting `x = -1`

`1 = C(-1)`

`C = -1`

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

Let `x = 1`:

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

`2-4 = 2B`

`B = -1`

The partial fraction expansion is:

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

Check:

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

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

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

This checks.

The original integrand is equal to the partial fractions:

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

Separate into 3 integrals:

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

These integrals are well known:

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