# How do you integrate (x+4)/((x+1)(x-2)^2 ) using partial fractions?

Nov 22, 2017

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

#### Explanation:

Perform the decomposition into partial fractions

$\frac{x + 4}{\left(x + 1\right) {\left(x - 2\right)}^{2}} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{x - 2} ^ 2$

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

The denominators are the same, compare the numerators

$x + 4 = A {\left(x - 2\right)}^{2} + B \left(x + 1\right) \left(x - 2\right) + C \left(x + 1\right)$

Let $x = - 1$, $\implies$, $3 = 9 A$, $\implies$, $A = \frac{1}{3}$

Let $x = 2$, $\implies$, $6 = 3 C$, $\implies$, $C = 2$

Coefficients of ${x}^{2}$

$0 = A + B$, $\implies$, $B = - \frac{1}{3}$

Therefore,

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

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

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