Question

How do you evaluate the integral `int 1/(x(x-1)^3)`?

Answer 1

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

Explanation:

Write:

`1/(x(x-1)^3) = A/x+B/(x-1)+C/(x-1)^2+D/(x-1)^3`

Multiplying both sides by `x(x-1)^3` we get:

`1 = A(x-1)^3+Bx(x-1)^2+Cx(x-1)+Dx`

Putting `x=0` we find `A=-1`

Looking at the coefficient of `x^3`, we find `B=-A = 1`

Putting `x=1` we find `D=1`

Looking at the coefficient of `x`, we find:

`0 = 3A+B-C+D`

Hence `C=-1`

So

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

and:

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

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