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

May 13, 2016

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

#### Explanation:

Let us first find partial fractions of $\frac{1}{\left(3 + x\right) \left(1 - x\right)}$ and for this let

$\frac{1}{\left(3 + x\right) \left(1 - x\right)} \Leftrightarrow \frac{A}{3 + x} + \frac{B}{1 - x}$ or

$\frac{1}{\left(3 + x\right) \left(1 - x\right)} \Leftrightarrow \frac{A \left(1 - x\right) + B \left(3 + x\right)}{\left(3 + x\right) \left(1 - x\right)}$ or

$\frac{1}{\left(3 + x\right) \left(1 - x\right)} \Leftrightarrow \frac{\left(B - A\right) x + \left(A + 3 B\right)}{\left(3 + x\right) \left(1 - x\right)}$ or

Hence $B - A = 0$ or $A = B$ and $A + 3 B = 1$, i.e.$A = B = \frac{1}{4}$

Hence $\int \frac{1}{\left(3 + x\right) \left(1 - x\right)} \mathrm{dx} = \int \left[\frac{1}{4 \left(3 + x\right)} + \frac{1}{4 \left(1 - x\right)}\right] \mathrm{dx}$ or

= $\frac{1}{4} \int \frac{1}{3 + x} \mathrm{dx} + \frac{1}{4} \int \frac{1}{1 - x} \mathrm{dx}$

= $\frac{1}{4} \ln \left(3 + x\right) - \frac{1}{4} \ln \left(1 - x\right)$