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

Dec 16, 2016

$= 2 \left(\ln | x - 1 | - \ln | x | + \frac{1}{x}\right) + C$

#### Explanation:

${x}^{3} - {x}^{2}$ can be factored as ${x}^{2} \left(x - 1\right)$.

$\frac{A x + B}{{x}^{2}} + \frac{C}{x - 1} = \frac{2}{\left({x}^{2}\right) \left(x - 1\right)}$

$\left(A x + B\right) \left(x - 1\right) + C \left({x}^{2}\right) = 2$

$A {x}^{2} + B x - A x - B + C {x}^{2} = 2$

$\left(A + C\right) {x}^{2} + \left(B - A\right) x + \left(- B\right) = 2$

We can now write a system of equations.

$\left\{\begin{matrix}A + C = 0 \\ B - A = 0 \\ - B = 2\end{matrix}\right.$

Solving, we obtain $B = - 2$, $A = - 2$, $C = 2$.

Therefore, the partial fraction decomposition is:

$\frac{- 2 x - 2}{x} ^ 2 + \frac{2}{x - 1}$

We now decompose $\frac{- 2 x - 2}{x} ^ 2$ into partial fractions.

$\frac{A}{x} ^ 2 + \frac{B}{x} = \frac{- 2 x - 2}{x} ^ 2$

$A + B x = - 2 x - 2 \to A = - 2 \mathmr{and} B = - 2$

Therefore, the complete partial fraction decomposition of $\frac{2}{{x}^{3} - {x}^{2}}$ is $- \frac{2}{x} ^ 2 - \frac{2}{x} + \frac{2}{x - 1}$.

This can be integrated as follows:

$\int \left(- \frac{2}{x} ^ 2 - \frac{2}{x} + \frac{2}{x - 1}\right) \mathrm{dx}$

$= - 2 \ln | x | + 2 \ln | x - 1 | - \int \left(2 {x}^{-} 2\right) + C$

$= 2 \ln | x - 1 | - 2 \ln | x | - \left(- 2 {x}^{-} 1\right) + C$

$= 2 \ln | x - 1 | - 2 \ln | x | + \frac{2}{x} + C$

$= 2 \left(\ln | x - 1 | - \ln | x | + \frac{1}{x}\right) + C$

Hopefully this helps!