# How do you integrate f(x)=(-x^2-2x)/((x^2+2)(x+7)) using partial fractions?

Sep 21, 2016

$- \frac{8}{51} \ln \left({x}^{2} + 2\right) + \frac{5 \sqrt{2}}{51} \arctan \left(\frac{x}{\sqrt{2}}\right) - \frac{35}{51} \ln \left\mid x + 7 \right\mid + C$

#### Explanation:

$\frac{- {x}^{2} - 2 x}{\left({x}^{2} + 2\right) \left(x + 7\right)} = \frac{A x + B}{{x}^{2} + 2} + \frac{C}{x + 7} =$

$= \frac{\left(A x + B\right) \left(x + 7\right)}{\left({x}^{2} + 2\right) \left(x + 7\right)} + \frac{C \left({x}^{2} + 2\right)}{\left({x}^{2} + 2\right) \left(x + 7\right)} =$

$= \frac{A {x}^{2} + 7 A x + B x + 7 B + C {x}^{2} + 2 C}{\left({x}^{2} + 2\right) \left(x + 7\right)} =$

$= \frac{\left(A + C\right) {x}^{2} + \left(7 A + B\right) x + 7 B + 2 C}{\left({x}^{2} + 2\right) \left(x + 7\right)}$

$\frac{- {x}^{2} - 2 x}{\left({x}^{2} + 2\right) \left(x + 7\right)} = \frac{\left(A + C\right) {x}^{2} + \left(7 A + B\right) x + 7 B + 2 C}{\left({x}^{2} + 2\right) \left(x + 7\right)}$

$A + C = - 1 \implies 7 A + 7 C = - 7$
$7 A + B = - 2$
$7 B + 2 C = 0$

$B - 7 C = 5 = > 7 B - 49 C = 35$
$7 B + 2 C = 0$

$51 C = - 35 \implies C = - \frac{35}{51}$

$B = 5 + 7 C = 5 - \frac{245}{51} = \frac{10}{51}$

$A = - 1 - C = - 1 + \frac{35}{51} = - \frac{16}{51}$

$I = \int \frac{- {x}^{2} - 2 x}{\left({x}^{2} + 2\right) \left(x + 7\right)} \mathrm{dx} = \int \frac{- \frac{16}{51} x + \frac{10}{51}}{{x}^{2} + 2} \mathrm{dx} + \int \frac{- \frac{35}{51}}{x + 7} \mathrm{dx}$

$I = - \frac{16}{51} \int \frac{x \mathrm{dx}}{{x}^{2} + 2} + \frac{10}{51} \int \frac{\mathrm{dx}}{{x}^{2} + 2} - \frac{35}{51} \int \frac{\mathrm{dx}}{x + 7}$

$I = - \frac{16}{51} \int \frac{\frac{1}{2} d \left({x}^{2} + 2\right)}{{x}^{2} + 2} + \frac{10}{51} \int \frac{\mathrm{dx}}{{x}^{2} + {\left(\sqrt{2}\right)}^{2}} - \frac{35}{51} \int \frac{\mathrm{dx}}{x + 7}$

$I = - \frac{8}{51} \ln \left({x}^{2} + 2\right) + \frac{5 \sqrt{2}}{51} \arctan \left(\frac{x}{\sqrt{2}}\right) - \frac{35}{51} \ln \left\mid x + 7 \right\mid + C$