# How do you integrate 1/(x^3-5x^2) using partial fractions?

May 13, 2016

$\int \frac{1}{{x}^{2} \left(x - 5\right)} \mathrm{dx} = - \frac{1}{25} \ln x + \frac{1}{5 x} + \frac{1}{25} \ln \left(x - 5\right)$

#### Explanation:

Let us first find partial fractions of 1/(x^3-5x^2)=1/(x^2(x-5) and for this let

$\frac{1}{{x}^{2} \left(x - 5\right)} \Leftrightarrow \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x - 5}$ or

$\frac{1}{{x}^{2} \left(x - 5\right)} \Leftrightarrow \frac{A x \left(x - 5\right) + B \left(x - 5\right) + C {x}^{2}}{{x}^{2} \left(x - 5\right)}$ or

$\frac{1}{{x}^{2} \left(x - 5\right)} \Leftrightarrow \frac{\left(A + C\right) {x}^{2} + \left(B - 5 A\right) x - 5 B}{{x}^{2} \left(x - 5\right)}$ or

Hence, $A + C = 0$, $B - 5 A = 0$ and $- 5 B = 1$ i.e.

$B = - \frac{1}{5}$, $A = \frac{1}{5} B = - \frac{1}{25}$ and $C = \frac{1}{25}$

Hence $\int \frac{1}{{x}^{2} \left(x - 5\right)} \mathrm{dx} = \int \left[- \frac{1}{25 x} - \frac{1}{5 {x}^{2}} + \frac{1}{25 \left(x - 5\right)}\right] \mathrm{dx}$

= $- \frac{1}{25} \int \frac{\mathrm{dx}}{x} - \frac{1}{5} \int \frac{\mathrm{dx}}{x} ^ 2 + \frac{1}{25} \int \frac{\mathrm{dx}}{x - 5}$

= $- \frac{1}{25} \ln x + \frac{1}{5 x} + \frac{1}{25} \ln \left(x - 5\right)$