# How do you integrate 10/(5x^2-2x^3) using partial fractions?

Feb 2, 2017

The answer is $= - \frac{2}{x} + \frac{4}{5} \ln \left(| x |\right) - \frac{4}{5} \ln \left(| 2 x - 5 |\right) + C$

#### Explanation:

$5 {x}^{2} - 2 {x}^{3} = - {x}^{2} \left(2 x - 5\right)$

Let's perform the decompostion into partial fractions

$\frac{10}{5 {x}^{2} - 2 {x}^{3}} = - \frac{10}{\left({x}^{2} \left(2 x - 5\right)\right)}$

$= \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{2 x - 5}$

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

The denpminators are the same, we can compare the numerators

$- 10 = A \left(2 x - 5\right) + B x \left(2 x - 5\right) + C {x}^{2}$

Let $x = 0$, $\implies$, $- 10 = - 5 A$, $\implies$, $A = 2$

Let $x = \frac{5}{2}$, $\implies$, $- 10 = \frac{25}{4} C$, $\implies$, $C = - \frac{8}{5}$

Coefficients of ${x}^{2}$

$0 = 2 B + C$, $\implies$, $B = - \frac{C}{2} = \frac{1}{2} \cdot \frac{8}{5} = \frac{4}{5}$

Therefore,

$\frac{10}{5 {x}^{2} - 2 {x}^{3}} = \frac{2}{x} ^ 2 + \frac{\frac{4}{5}}{x} + \frac{- \frac{8}{5}}{2 x - 5}$

So,

$\int \frac{10 \mathrm{dx}}{5 {x}^{2} - 2 {x}^{3}} = 2 \int \frac{\mathrm{dx}}{x} ^ 2 + \frac{4}{5} \int \frac{\mathrm{dx}}{x} - \frac{8}{5} \int \frac{\mathrm{dx}}{2 x - 5}$

$= - \frac{2}{x} + \frac{4}{5} \ln \left(| x |\right) - \frac{4}{5} \ln \left(| 2 x - 5 |\right) + C$