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

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Answer 1 · 2018-06-10T21:14:33

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

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:

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

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