# How do you integrate int (x^2)/(x+1)^3 dx using partial fractions?

Nov 3, 2016

$\int \textcolor{w h i t e}{.} {x}^{2} / {\left(x + 1\right)}^{3} \textcolor{w h i t e}{.} \mathrm{dx} = \ln \left\mid x + 1 \right\mid + \frac{2}{x + 1} - \frac{1}{2 {\left(x + 1\right)}^{2}} + C$

#### Explanation:

$\int \textcolor{w h i t e}{.} {x}^{2} / {\left(x + 1\right)}^{3} \textcolor{w h i t e}{.} \mathrm{dx} = \int \textcolor{w h i t e}{.} \frac{\left({x}^{2} + 2 x + 1\right) - \left(2 x + 2\right) + 1}{x + 1} ^ 3 \textcolor{w h i t e}{.} \mathrm{dx}$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} {x}^{2} / {\left(x + 1\right)}^{3} \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \frac{{\left(x + 1\right)}^{2} - 2 \left(x + 1\right) + 1}{x + 1} ^ 3 \textcolor{w h i t e}{.} \mathrm{dx}$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} {x}^{2} / {\left(x + 1\right)}^{3} \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \frac{1}{x + 1} - \frac{2}{x + 1} ^ 2 + \frac{1}{x + 1} ^ 3 \textcolor{w h i t e}{.} \mathrm{dx}$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} {x}^{2} / {\left(x + 1\right)}^{3} \textcolor{w h i t e}{.} \mathrm{dx}} = \ln \left\mid x + 1 \right\mid + \frac{2}{x + 1} - \frac{1}{2 {\left(x + 1\right)}^{2}} + C$

Nov 3, 2016

The answer is $= - \frac{1}{2 {\left(x + 1\right)}^{2}} + \frac{2}{x + 1} + \ln \left(x + 1\right) + C$

#### Explanation:

Let's do the decomposition in partial fractions
${x}^{2} / {\left(x + 1\right)}^{3} = \frac{A}{x + 1} ^ 3 + \frac{B}{x + 1} ^ 2 + \frac{C}{x + 1}$
$= \frac{A + B \left(x + 1\right) + C {\left(x + 1\right)}^{2}}{x + 1} ^ 3$
We need to find A,B,C
So ${x}^{2} = A + B \left(x + 1\right) + C {\left(x + 1\right)}^{2}$
coefficients of ${x}^{2}$$\implies$$C = 1$
let $x = - 1$$\implies$$A = 1$
Coefficients of x, $0 = B + 2 C$$\implies$$B = - 2$
$\therefore {x}^{2} / {\left(x + 1\right)}^{3} = \frac{1}{x + 1} ^ 3 - \frac{2}{x + 1} ^ 2 + \frac{1}{x + 1}$
So $\int \frac{{x}^{2} \mathrm{dx}}{x + 1} ^ 3 = \int \frac{\mathrm{dx}}{x + 1} ^ 3 - \int \frac{2 \mathrm{dx}}{x + 1} ^ 2 + \int \frac{\mathrm{dx}}{x + 1}$
We use $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$ and $\int \frac{\mathrm{dx}}{x} = \ln x$
$\int \frac{{x}^{2} \mathrm{dx}}{x + 1} ^ 3 = - \frac{1}{2 {\left(x + 1\right)}^{2}} + \frac{2}{x + 1} + \ln \left(x + 1\right) + C$