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

Sep 14, 2016

$\ln | x - 1 | + 2 \ln | x + 4 | + C$,

or,

$\ln | \left(x - 1\right) {\left(x + 4\right)}^{2} | + C$.

#### Explanation:

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

To decompose the integrand using Method of Partial Fraction, let,

$\frac{3 x + 2}{\left(x - 1\right) \left(x + 4\right)} = \frac{A}{x - 1} + \frac{B}{x + 4} \text{, where, } A , B \in \mathbb{R}$.

We use Heavyside's Cover-up Method to determine $A , \mathmr{and} , B$ :

$A = {\left[\frac{3 x + 2}{x + 4}\right]}_{x = 1} = \frac{3 + 2}{1 + 4} = 1$

$B = {\left[\frac{3 x + 2}{x - 1}\right]}_{x = - 4} = \frac{- 12 + 2}{- 4 - 1} = 2$. Hence,

$I = \int \left[\frac{1}{x - 1} + \frac{2}{x + 4}\right] \mathrm{dx}$

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

$= \ln | x - 1 | + 2 \ln | x + 4 |$

Therefore,

$I = \ln | x - 1 | + 2 \ln | x + 4 + C |$, or,

$I = \ln | \left(x - 1\right) {\left(x + 4\right)}^{2} | + C$.

Enjoy Maths.!