# How do you find int (x-1) / ((2x+1) (4x+4)) dx using partial fractions?

Oct 22, 2016

#### Explanation:

Please notice that the integral can be multiplied by $\frac{1}{4}$ and the integrand becomes:

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

Multiply by the left denominator:

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

Let $x = - 1$

$- 2 = - B$

$B = 2$

Let (x = -1/2)

$- \frac{3}{2} = A \left(\frac{1}{2}\right)$

$A = - 3$

Check:

$\frac{2}{x + 1} \frac{2 x + 1}{2 x + 1} - \frac{3}{2 x + 1} \frac{x + 1}{x + 1} =$

$\frac{4 x + 2 - 3 x - 3}{\left(2 x + 1\right) \left(x + 1\right)} =$

$\frac{x - 1}{\left(2 x + 1\right) \left(x + 1\right)}$ This checks:

$\int \frac{x - 1}{\left(2 x + 1\right) \left(4 x + 4\right)} \mathrm{dx} = \frac{1}{2} \int \frac{1}{x + 1} \mathrm{dx} - \frac{3}{4} \int \frac{1}{2 x + 1} \mathrm{dx}$

$\int \frac{x - 1}{\left(2 x + 1\right) \left(4 x + 4\right)} \mathrm{dx} = \frac{1}{2} \ln | x + 1 | - \frac{3}{8} \ln | 2 x + 1 | + C$