# How do you find int 8/ ( x^2 (x+3) )dx using partial fractions?

Dec 4, 2015

Apply partial fractions to find
$\int \frac{8}{{x}^{2} \left(x + 3\right)} \mathrm{dx} = \frac{8}{9} \left(\ln | \frac{x + 3}{x} | - \frac{3}{x}\right) + C$

#### Explanation:

In partial fraction decomposition, we consider denominators which could combine to create the original denominator as the common denominator, and then solve for the numerators.

As the factors of ${x}^{2} \left(x + 3\right)$ are $x$, ${x}^{2}$, and $x + 3$ we proceed as follows:

$\frac{8}{{x}^{2} \left(x + 3\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 3}$

Now, multiplying through by ${x}^{2} \left(x + 3\right)$ we obtain

$8 = A x \left(x + 3\right) + B \left(x + 3\right) + C {x}^{2}$

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

Now, we equate corresponding coefficients to obtain the system
$\left\{\begin{matrix}A + C = 0 \\ 3 A + B = 0 \\ 3 B = 8\end{matrix}\right.$

Solving the system, we obtain
$\left\{\begin{matrix}A = - \frac{8}{9} \\ B = \frac{8}{3} \\ C = \frac{8}{9}\end{matrix}\right.$

Then we substitute back into our original equation to obtain

$\frac{8}{{x}^{2} \left(x + 3\right)} = \frac{- \frac{8}{9}}{x} + \frac{\frac{8}{3}}{x} ^ 2 + \frac{\frac{8}{9}}{x + 3}$

$\implies \frac{8}{{x}^{2} \left(x + 3\right)} = - \frac{8}{9 x} + \frac{8}{3 {x}^{2}} + \frac{8}{9 \left(x + 3\right)}$

Now we proceed to integrate.

$\int \frac{8}{{x}^{2} \left(x + 3\right)} \mathrm{dx} = \int \left(- \frac{8}{9 x} + \frac{8}{3 {x}^{2}} + \frac{8}{9 \left(x + 3\right)}\right) \mathrm{dx}$

$= \int - \frac{8}{9 x} \mathrm{dx} + \int \frac{8}{3 {x}^{2}} \mathrm{dx} + \int \frac{8}{9 \left(x + 3\right)} \mathrm{dx}$

$= - \frac{8}{9} \int \frac{1}{x} \mathrm{dx} + \frac{8}{3} \int \frac{1}{x} ^ 2 \mathrm{dx} + \frac{8}{9} \int \frac{1}{x + 3} \mathrm{dx}$

$= - \frac{8}{9} \ln | x | + \frac{8}{3} \left(- \frac{1}{x}\right) + \frac{8}{9} \ln | x + 3 | + C$

$= \frac{8}{9} \left(\ln | \frac{x + 3}{x} | - \frac{3}{x}\right) + C$