How do you integrate int 18/ ( x^2 (x+3) ) dx using partial fractions?

Feb 9, 2016

$\int \frac{18}{{x}^{2} \left(x + 3\right)} \mathrm{dx} = - 2 \ln | x | - \frac{6}{x} + 2 \ln | x + 3 | + C$

Explanation:

$\frac{18}{{x}^{2} \left(x + 3\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 3}$

$\implies 18 = A x \left(x + 3\right) + B \left(x + 3\right) + C {x}^{2}$

$= \left(A + C\right) {x}^{2} + \left(3 A + B\right) x + 3 B$

$\implies \left\{\begin{matrix}A + C = 0 \\ 3 A + B = 0 \\ 3 B = 18\end{matrix}\right.$

$\implies \left\{\begin{matrix}A = - 2 \\ B = 6 \\ C = 2\end{matrix}\right.$

$\implies \frac{18}{{x}^{2} \left(x + 3\right)} = - \frac{2}{x} + \frac{6}{x} ^ 2 + \frac{2}{x + 3}$

$\implies \int \frac{18}{{x}^{2} \left(x + 3\right)} \mathrm{dx} = \int - \frac{2}{x} + \frac{6}{x} ^ 2 + \frac{2}{x + 3} \mathrm{dx}$

$= - 2 \int \frac{1}{x} \mathrm{dx} + 6 \int \frac{1}{x} ^ 2 \mathrm{dx} + 2 \int \frac{1}{x + 3} \mathrm{dx}$

$= - 2 \ln | x | - \frac{6}{x} + 2 \ln | x + 3 | + C$