# How do you integrate int (x^3)/(x^2 + 8x + 16) using partial fractions?

Feb 1, 2017

$\frac{1}{2} {x}^{2} - 8 x + 48 \ln | x + 4 | + \frac{64}{x + 4} + c$

#### Explanation:

The denominator is a perfect square, so you don't use partial fractions. Instead, you substitute $u = x + 4$. The integral becomes:
$\int {\left(u - 4\right)}^{3} / {u}^{2} \mathrm{du}$

$= \int \frac{{u}^{3} - 12 {u}^{2} + 48 u - 64}{u} ^ 2 \mathrm{du}$

$= \int u - 12 + \frac{48}{u} - \frac{64}{u} ^ 2 \mathrm{du}$

$= \frac{1}{2} {u}^{2} - 12 u + 48 \ln | u | + \frac{64}{u} + k$

$= \frac{1}{2} {\left(x + 4\right)}^{2} - 12 \left(x + 4\right) + 48 \ln | x + 4 | + \frac{64}{x + 4} + k$

$= \frac{1}{2} {x}^{2} - 8 x + 48 \ln | x + 4 | + \frac{64}{x + 4} + c$

(where $c = 8 - 48 + k$)