# How do you evaluate the integral int dx/((2x-1)(x+2))?

Jan 5, 2017

Use partial fractions. See below.

$\int \frac{1}{\left(2 x - 1\right) \left(x + 2\right)} \mathrm{dx} = \frac{1}{5} \ln \left(| 2 x - 1 |\right) - \frac{1}{5} \ln \left(| x + 2 |\right) + C$

#### Explanation:

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

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

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

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

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

$x = - 2$

$1 = B \left(- 4 - 1\right)$

$1 = - 5 B$

$B = - \frac{1}{5}$

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

$1 = A \left(\frac{1}{2} + 2\right)$

$1 = A \left(\frac{5}{2}\right)$

$A = \frac{2}{5}$

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

$\implies \int \frac{\frac{2}{5}}{2 x - 1} + \frac{- \frac{1}{5}}{x + 2} \mathrm{dx}$

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

$\implies \frac{2}{5} \int \frac{1}{2 x - 1} \mathrm{dx} - \frac{1}{5} \int \frac{1}{x + 2} \mathrm{dx}$

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

For the second integral, $z = x + 2 , \mathrm{dz} = \mathrm{dx}$.

$\implies \frac{1}{5} \int \frac{1}{u} \mathrm{du} - \frac{1}{5} \int \frac{1}{z} \mathrm{dz}$

Integrate.

$\implies \frac{1}{5} \ln | u | - \frac{1}{5} \ln | z | + C$

Substitute back in.

$\implies \frac{1}{5} \ln | 2 x - 1 | - \frac{1}{5} \ln | x + 2 | + C$

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

By rules of logarithms, this is equivalent to

$\frac{1}{5} \ln \left(| \frac{2 x - 1}{x + 2} |\right) + C$