# Integral of int(x-1)/(x+3)^3 dx ?

## $\int \frac{x - 1}{x + 3} ^ 3 \mathrm{dx}$

Apr 14, 2018

$- \frac{1}{x + 3} + \frac{2}{x + 3} ^ 2 + c$

#### Explanation:

First, just look at $\frac{x - 1}{x + 3} ^ 3$

$\frac{x - 1}{x + 3} ^ 3 = \frac{a}{x + 3} + \frac{b}{x + 3} ^ 2 + \frac{c}{x + 3} ^ 3 | \cdot {\left(x + 3\right)}^{3}$
$x - 1 = a \cdot {\left(x + 3\right)}^{2} + b \cdot \left(x + 3\right) + c$
$\textcolor{red}{0} {x}^{2} + \textcolor{b l u e}{1} x \textcolor{\mathmr{and} a n \ge}{- 1} = \left(a\right) {x}^{2} + \left(6 a + b\right) x + 9 a + 3 b + c$

$| \left(a = \textcolor{red}{0}\right) , \left(6 a + b = \textcolor{b l u e}{1}\right) , \left(9 a + 3 b + c = \textcolor{\mathmr{and} a n \ge}{- 1}\right) |$

$| \left(a = 0\right) , \left(b = 1\right) , \left(c = - 4\right) |$

$\frac{x - 1}{x + 3} ^ 3 = \frac{0}{x + 3} + \frac{1}{x + 3} ^ 2 - \frac{4}{x + 3} ^ 3$
$\frac{x - 1}{x + 3} ^ 3 = \frac{1}{x + 3} ^ 2 - \frac{4}{x + 3} ^ 3$

$\int \frac{x - 1}{x + 3} ^ 3 \mathrm{dx} = \int \left(\frac{1}{x + 3} ^ 2 - \frac{4}{x + 3} ^ 3\right) \mathrm{dx}$
$\int \left(\frac{1}{x + 3} ^ 2 - \frac{4}{x + 3} ^ 3\right) \mathrm{dx} = - \frac{1}{x + 3} + \frac{2}{x + 3} ^ 2 + c$