# How do you use partial fraction decomposition to decompose the fraction to integrate 1/((x+2)(x^2+4))?

Oct 24, 2015

$I = \frac{1}{8} \ln | x + 2 | - \frac{1}{16} \ln \left({x}^{2} + 4\right) + \frac{1}{8} \arctan \left(\frac{x}{2}\right) + C$

#### Explanation:

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

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

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

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

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

$A + B = 0 \implies A = - B$
$2 B + C = 0 \implies C = - 2 B$
$4 A + 2 C = 1 \implies 4 \left(- B\right) + 2 \left(- 2 B\right) = 1$

$4 \left(- B\right) + 2 \left(- 2 B\right) = 1$

$- 8 B = 1 \implies B = - \frac{1}{8}$

$A = - B \implies A = \frac{1}{8}$

$C = - 2 B \implies C = \frac{1}{4}$

$I = \int \frac{1}{\left(x + 2\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \frac{1}{8} \int \frac{1}{x + 2} \mathrm{dx} + \int \frac{- \frac{1}{8} x + \frac{1}{4}}{{x}^{2} + 4} \mathrm{dx}$

$I = \frac{1}{8} \ln | x + 2 | - \frac{1}{8} \int \frac{x \mathrm{dx}}{{x}^{2} + 4} + \frac{1}{4} \int \frac{\mathrm{dx}}{{x}^{2} + 4}$

$I = \frac{1}{8} \ln | x + 2 | - \frac{1}{16} \int \frac{\left({x}^{2} + 4\right) ' \mathrm{dx}}{{x}^{2} + 4} + \frac{1}{4} \cdot \frac{1}{2} \arctan \left(\frac{x}{2}\right)$

$I = \frac{1}{8} \ln | x + 2 | - \frac{1}{16} \int \frac{d \left({x}^{2} + 4\right)}{{x}^{2} + 4} + \frac{1}{8} \arctan \left(\frac{x}{2}\right)$

$I = \frac{1}{8} \ln | x + 2 | - \frac{1}{16} \ln \left({x}^{2} + 4\right) + \frac{1}{8} \arctan \left(\frac{x}{2}\right) + C$