# How do you integrate int (4x - 16) / (x^2 - 2x - 3) using partial fractions?

## How do you integrate $\int \frac{4 x - 16}{{x}^{2} - 2 x - 3} \mathrm{dx}$ using partial fractions?

Sep 4, 2016

$- \ln | x - 3 | + 5 \ln | x + 1 | + C$,

OR,

$\ln | {\left(x + 1\right)}^{5} / \left(x - 3\right) | + C$.

#### Explanation:

Let $I = \int \frac{4 x - 16}{{x}^{2} - 2 x - 3} \mathrm{dx} = 4 \int \frac{x - 4}{\left(x - 3\right) \left(x + 1\right)} \mathrm{dx}$

We decompose the Integrand

$\frac{x - 4}{\left(x - 3\right) \left(x + 1\right)} = \frac{A}{x - 3} + \frac{B}{x + 1} , \text{ where, } A , B \in \mathbb{R}$.

Using Heavyside's Cover-up Method to find consts. $A , B \in \mathbb{R}$,

$A = {\left[\frac{x - 4}{x + 1}\right]}_{x = 3} = \frac{3 - 4}{3 + 1} = - \frac{1}{4}$,

$B = {\left[\frac{x - 4}{x - 3}\right]}_{x = - 1} = \frac{- 1 - 4}{- 1 - 3} = \frac{5}{4}$.

Hence, $I = 4 \int \left[\frac{- \frac{1}{4}}{x - 3} + \frac{\frac{5}{4}}{x + 1}\right] \mathrm{dx}$

$= - \int \frac{1}{x - 3} \mathrm{dx} + 5 \int \frac{1}{x + 1} \mathrm{dx}$

$\therefore I = - \ln | x - 3 | + 5 \ln | x + 1 | + C$, or,

$I = \ln | {\left(x + 1\right)}^{5} / \left(x - 3\right) | + C$.

Enjoy Maths.!