Question

How do you integrate `int x/ (x^3-2x^2+x)` using partial fractions?

Answer 1

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

Explanation:

Simplify the integrand:

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

Factorize the denominator:

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

Then:

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