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

May 13, 2016

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

#### Explanation:

int (x+2) /(2x^2-x-1) * d x=?

$\int \frac{x + 2}{\left(2 x + 1\right) \cdot \left(x - 1\right)} d x$

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

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

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

$3 = 0 + 3 B \text{ } B = 1$

$x = - \frac{1}{2} \rightarrow \text{ } - \frac{1}{2} + 2 = A \left(- \frac{1}{2} - 1\right) + 0$

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

$\int \frac{x + 2}{2 {x}^{2} - x - 1} \cdot d x = - \int \frac{d x}{2 x + 1} + \int \frac{d x}{x - 1}$

$u = 2 x + 1 \text{ "d u=2d x ,d x=(d u)/2" ; "v=x-1" } d v = d x$

$\int \frac{x + 2}{2 {x}^{2} - x - 1} \cdot d x = - \frac{1}{2} \int \frac{d u}{u} + \int \frac{d v}{v}$

$\int \frac{x + 2}{2 {x}^{2} - x - 1} \cdot d x = - \frac{1}{2} l n u + l n v + C$

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