How do you integrate int (x+13)/(x^3+2x^2-5x-6) using partial fractions?

Oct 30, 2016

The answer is $\ln \left(x - 2\right) - 2 \ln \left(x + 1\right) + \ln \left(x + 3\right) + C$

Explanation:

Let's start by factorising the denominator
By trial and error we finf that $\left(x - 2\right)$ is a factor
Let's do a long division
${x}^{3} + 2 {x}^{2} - 5 x - 6$$\textcolor{w h i t e}{a a a a}$∣$x - 2$
${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$∣${x}^{2} + 4 x + 3$
$0 + 4 {x}^{2} - 5 x$
$\textcolor{w h i t e}{a a a}$$4 {x}^{2} - 8 x$
$\textcolor{w h i t e}{a a a a a}$$0 + 3 x - 6$
$\textcolor{w h i t e}{a a a a a}$$0 + 3 x - 6$
$\textcolor{w h i t e}{a a a a a a a a a}$$0 - 0$

So the factorisation is ${x}^{2} + 4 x + 3 = \left(x + 1\right) \left(x + 3\right)$
And ${x}^{3} + 2 {x}^{2} - 5 x - 6 = \left(x - 2\right) \left(x + 1\right) \left(x + 3\right)$
So the partial fractiona are
$\frac{x + 13}{\left(x - 2\right) \left(x + 1\right) \left(x + 3\right)} = \frac{A}{x - 2} + \frac{B}{x + 1} + \frac{C}{x + 3}$
$= \frac{A \left(x + 1\right) \left(x + 3\right) + B \left(x - 2\right) \left(x + 3\right) + C \left(x - 2\right) \left(x + 1\right)}{\left(x - 2\right) \left(x + 1\right) \left(x + 3\right)}$
(x+13)=A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1))
let $x = 2$$\implies$$15 = 15 A$$\implies$$A = 1$
let $x = - 1$$\implies$$12 = - 6 B$$\implies$$B = - 2$
let$x = - 3$$\implies$$10 = 10 C$$\implies$$C = 1$
So the integration is
$\int \frac{\left(x + 13\right) \mathrm{dx}}{{x}^{3} + 2 {x}^{2} - 5 x - 6} = \int \frac{\mathrm{dx}}{x - 2} - \int \frac{2 \mathrm{dx}}{x + 1} + \int \frac{\mathrm{dx}}{x + 3}$
$\ln \left(x - 2\right) - 2 \ln \left(x + 1\right) + \ln \left(x + 3\right) + C$