# How do you integrate int 5/ ( (2x-3) (x+1) ) dx using partial fractions?

Mar 11, 2016

$\int \frac{5}{\left(2 x - 3\right) \left(x + 1\right)} d x = l n \left(2 x - 3\right) - l n \left(x + 1\right) + C$

#### Explanation:

$\int \frac{5}{\left(2 x - 3\right) \left(x + 1\right)} d x = 5 \int \frac{d x}{\left(2 x - 3\right) \left(x + 1\right)}$
$\frac{1}{\left(2 x - 3\right) \left(x + 1\right)} = \frac{A}{2 x - 3} + \frac{B}{x + 1}$
$\frac{1}{\left(2 x - 3\right) \left(x + 1\right)} = \frac{A \left(x + 1\right) + B \left(2 x - 3\right)}{\left(2 x - 3\right) \left(x + 1\right)}$
$\text{for x="-1 =>" " 1=B(2*(-1)-3)" } B = - \frac{1}{5}$
$\text{for x="3/2 =>" " 1=A(3/2+1)" } A = \frac{2}{5}$
$5 \int \frac{d x}{\left(2 x - 3\right) \left(x + 1\right)} = 5 \left(\int \frac{2 d x}{5 \left(2 x - 3\right)} - \int \frac{d x}{5 \left(x + 1\right)}\right)$
$= \int \frac{2 d x}{\left(2 x - 3\right)} - \int \frac{d x}{\left(x + 1\right)}$
$u = 2 x - 3 \text{ "d u=2d x" "v=x+1" } d v = d x$
$= \int \frac{d u}{u} - \int \frac{d v}{v} + C$
$\int \frac{5}{\left(2 x - 3\right) \left(x + 1\right)} d x = l n u - l n v + C$
$\int \frac{5}{\left(2 x - 3\right) \left(x + 1\right)} d x = l n \left(2 x - 3\right) - l n \left(x + 1\right) + C$