How do you evaluate the integral of int (2-x)/(1- x^2)?

Jan 2, 2016

$- \frac{1}{2} \ln | 1 - x | + \frac{3}{2} \ln | 1 + x | + C$ or

$\frac{1}{2} \left(3 \ln | 1 + x | - \ln | 1 - x |\right) + C$

Explanation:

Use partial fraction decomposition to rewrite the integrant before evaluate the integral

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

As partial fraction
$\frac{A}{1 - x} + \frac{B}{1 + x} = \frac{2 - x}{\left(1 - x\right) \left(1 + x\right)}$

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

$x : \text{ " } A - B = - 1$
${x}^{0} \text{ " } A + B = 2$

When you solve the system

A= 1/2 ; " " " B= 3/2

Rewrite the integral as

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

$- \frac{1}{2} \ln | 1 - x | + \frac{3}{2} \ln | 1 + x | + C$ or

$\frac{1}{2} \left(3 \ln | 1 + x | - \ln | 1 - x |\right) + C$