# How do you use partial fractions to find the integral int (5x^2-12x-12)/(x^3-4x)dx?

Nov 4, 2017

$\int \frac{\left(5 {x}^{2} - 12 x - 12\right) \cdot \mathrm{dx}}{{x}^{3} - 4 x}$=$4 L n \left(x + 2\right) + 3 L n x - 2 L n \left(x - 2\right) + C$

#### Explanation:

$\frac{5 {x}^{2} - 12 x - 12}{{x}^{3} - 4 x}$

=$\frac{5 {x}^{2} - 12 x - 12}{x \cdot \left(x + 2\right) \cdot \left(x - 2\right)}$

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

After expanding denominator,

$A \cdot x \cdot \left(x - 2\right) + B \cdot \left(x + 2\right) \cdot \left(x - 2\right) + C \cdot x \cdot \left(x + 2\right) = 5 {x}^{2} - 12 x - 12$

Set $x = - 2$, $8 A = 32$, so $A = 4$

Set $x = 0$, $- 4 B = - 12$, so $B = 3$

Set $x = 2$, $- 8 C = - 16$, so $C = - 2$

Thus,

$\int \frac{\left(5 {x}^{2} - 12 x - 12\right) \cdot \mathrm{dx}}{{x}^{3} - 4 x}$

=$\int \frac{4 \mathrm{dx}}{x + 2}$+$\int \frac{3 \mathrm{dx}}{x}$-$\int \frac{2 \mathrm{dx}}{x - 2}$

=$4 L n \left(x + 2\right) + 3 L n x - 2 L n \left(x - 2\right) + C$