# How do you integrate f(x)=(x^2-2)/((x+4)(x-2)(x-2)) using partial fractions?

Feb 16, 2017

The answer is $= \frac{7}{18} \ln \left(| x + 4 |\right) + \frac{11}{18} \ln \left(| x - 2 |\right) - \frac{1}{3 \left(x - 2\right)} + C$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{{x}^{2} - 2}{\left(x + 4\right) {\left(x - 2\right)}^{2}} = \frac{A}{x + 4} + \frac{B}{x - 2} ^ 2 + \frac{C}{x - 2}$

$= \frac{A {\left(x - 2\right)}^{2} + B \left(x + 4\right) + C \left(x - 2\right) \left(x + 4\right)}{\left(x + 4\right) {\left(x - 2\right)}^{2}}$

The denominators are the same, we compare the numerators

${x}^{2} - 2 = A {\left(x - 2\right)}^{2} + B \left(x + 4\right) + C \left(x - 2\right) \left(x + 4\right)$

Let $x = - 4$, $\implies$, $14 = 36 A$, $\implies$, $A = \frac{7}{18}$

Let $x = 2$, $\implies$, $2 = 6 B$, $\implies$, $B = \frac{1}{3}$

Coefficients of ${x}^{2}$,

$1 = A + C$, $\implies$, $C = 1 - A = 1 - \frac{7}{18} = \frac{11}{18}$

Therefore,

$\frac{{x}^{2} - 2}{\left(x + 4\right) {\left(x - 2\right)}^{2}} = \frac{\frac{7}{18}}{x + 4} + \frac{\frac{1}{3}}{x - 2} ^ 2 + \frac{\frac{11}{18}}{x - 2}$

$\int \frac{\left({x}^{2} - 2\right) \mathrm{dx}}{\left(x + 4\right) {\left(x - 2\right)}^{2}} = \frac{7}{18} \int \frac{\mathrm{dx}}{x + 4} + \frac{1}{3} \int \frac{\mathrm{dx}}{x - 2} ^ 2 + \frac{11}{18} \int \frac{\mathrm{dx}}{x - 2}$

$= \frac{7}{18} \ln \left(| x + 4 |\right) + \frac{11}{18} \ln \left(| x - 2 |\right) - \frac{1}{3 \left(x - 2\right)} + C$