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

Dec 3, 2015

$\int \frac{{x}^{2}}{{x}^{2} + 3 x + 2} \mathrm{dx} = x - 4 \ln | x + 2 | + \ln | x + 1 | + C$

#### Explanation:

Original
$\int {x}^{2} / \left({x}^{2} + 3 x + 2\right) \mathrm{dx} \text{ " " " " } \left(1\right)$

Step 1: Rewrite the expression as a proper fraction (using long division)

$\left(\frac{{x}^{2}}{{x}^{2} + 3 x + 2}\right) = 1 - \frac{3 x + 2}{{x}^{2} + 3 x + 2} \text{ " " " } \left(2\right)$

$\int 1 \mathrm{dx} - \int \left(\frac{3 x + 2}{{x}^{2} + 3 x + 2}\right) \mathrm{dx} \text{ " " " " " } \left(3\right)$

Step 2: Factor the denominator $\left(3\right)$

$\frac{3 x + 2}{{x}^{2} + 3 x + 2} = \frac{3 x + 2}{\left(x + 1\right) \left(x + 2\right)}$

Set up the partial fraction decomposition
$\frac{3 x + 2}{{x}^{2} + 3 x + 2} =$

$= \frac{3 x + 2}{\left(x + 1\right) \left(x + 2\right)} = \frac{\textcolor{red}{A}}{x + 2} + \frac{\textcolor{red}{B}}{x + 1} \text{ " " " " } \left(4\right)$

Multiply LCD (least common denominator)

$\left(3 x + 2\right) = A \left(x + 1\right) + B \left(x + 2\right)$

Let $x = - 1$
$\left(3 \cdot - 1 + 2\right) = A \left(- 1 + 1\right) + B \left(- 1 + 2\right)$
$\textcolor{red}{- 1 = B}$

Let $x = - 2$
$3 \cdot - 2 + 2 = A \left(- 2 + 1\right) + B \left(- 2 + 2\right)$
$- 4 = - A$
$\textcolor{red}{4 = A}$

Rewrite $\left(3\right)$
$\frac{\textcolor{red}{4}}{x + 2} - \frac{\textcolor{red}{1}}{x + 1}$

Rewrite (2)

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

$= \int 1 \cdot \mathrm{dx} - \int \frac{4}{x + 2} \mathrm{dx} + \int \frac{1}{x + 1} \mathrm{dx}$

$\textcolor{b l u e}{= x - 4 \ln | x + 2 | - \ln | x + 1 | + C} \text{ " " " } \left(5\right)$

I hope this help.