# How do you find int 3/((1 + x)(1 - 2x))dx using partial fractions?

Nov 15, 2015

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

#### Explanation:

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

Expanding the Right hand side, we get
(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x)
Equating, we get
(A * (1 - 2x) + B * (1 + x)) / ((1 + x) * (1 - 2x) = $\frac{3}{\left(1 + x\right) \cdot \left(1 - 2 x\right)}$

ie $A \cdot \left(1 - 2 x\right) + B \cdot \left(1 + x\right) = 3$
or $A - 2 A x + B + B x = 3$
or $\left(A + B\right) + x \cdot \left(- 2 A + B\right) = 3$
equating the coefficient of x to 0 and equating constants, we get

$A + B = 3$ and
$- 2 A + B = 0$
Solving for A & B, we get
$A = 1 \mathmr{and} B = 2$
Substituting in the integration, we get
$\int \frac{3}{\left(1 + x\right) \cdot \left(1 - 2 x\right)} \mathrm{dx}$ = int (1 /(1 + x) + 2 / (1 - 2x)) dx

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

= $\ln \left(1 + x\right) + 2 \cdot \ln \left(1 - 2 x\right) \cdot \left(- \frac{1}{2}\right)$

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

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