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

Jan 6, 2017

See below.

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

Explanation:

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

Multiply through by the denominator of the left-hand side:

$\left(1 + x\right) \left(1 - 2 x\right) \left[\frac{3}{\cancel{\left(1 + x\right) \left(1 - 2 x\right)}} = \frac{A}{\cancel{1 + x}} + \frac{B}{\cancel{1 - 2 x}}\right]$

=>$3 = A \left(1 - 2 x\right) + B \left(1 + x\right)$

The next step is to solve for $A$ and $B$. One way to do this is to pick values for $x$ which will cancel each variable.

$x = \frac{1}{2}$

$3 = \cancel{A \cdot 0} + \frac{3}{2} B$

$B = 2$

$x = - 1$

$3 = A \left(1 - \left(- 2\right)\right) + \cancel{B \cdot 0}$

$3 = 3 A$

$A = 1$

We put these values back into our partial fractions and replace as the integrand.

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

Technically, you should use a substitution before integrating. Split up the integral.

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

For the first integral, $u = 1 + x , \mathrm{du} = \mathrm{dx}$
For the second integral, $z = 1 - 2 x , \mathrm{dz} = - 2 \mathrm{dx} \implies - \frac{1}{2} \mathrm{dz} = \mathrm{dx}$

$\int \frac{1}{u} \mathrm{du} - \int \frac{1}{z} \mathrm{dz}$

Integrate.

$\ln \left(| u |\right) - \ln \left(| z |\right) + C$

Substitute back in.

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

Note: the absolute value signs account for the domain of the natural log function ($x > 0$).

By the properties of logarithms, you may also write the final answer as:

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