# How do you evaluate the integral int (x+2)/(x^2-2x-3)?

Feb 1, 2017

The answer is $= - \frac{1}{4} \ln \left(| x + 1 |\right) + \frac{5}{4} \ln \left(| x - 3 |\right) + C$

#### Explanation:

Let's factorise the denominator

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

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we can equalize the numerators

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

Let $x = - 1$, $\implies$, $1 = - 4 A$, $\implies$, $A = - \frac{1}{4}$

Let $x = 3$, $\implies$, $5 = 4 B$, $B = \frac{5}{4}$

Therefore,

$\frac{x + 2}{\left(x + 1\right) \left(x - 3\right)} = \frac{- \frac{1}{4}}{x + 1} + \frac{\frac{5}{4}}{x - 3}$

$\int \frac{\left(x + 2\right) \mathrm{dx}}{\left(x + 1\right) \left(x - 3\right)} = - \frac{1}{4} \int \frac{\mathrm{dx}}{x + 1} + \frac{5}{4} \int \frac{\mathrm{dx}}{x - 3}$

$= - \frac{1}{4} \ln \left(| x + 1 |\right) + \frac{5}{4} \ln \left(| x - 3 |\right) + C$