# How do you find int (11-2x)/(x^2 + x - 2) dx using partial fractions?

Oct 22, 2016

$\int \frac{\left(11 - 2 x\right) \mathrm{dx}}{\left(x - 1\right) \left(x + 2\right)} = 3 \ln \left(x - 1\right) - 5 \ln \left(x + 2\right) + C$

#### Explanation:

First factorise the denominator
${x}^{2} + x - 2 = \left(x - 1\right) \left(x + 2\right)$

Then we look for the partial fraction

$\frac{11 - 2 x}{\left(x - 1\right) \left(x + 2\right)} = \frac{A}{x - 1} + \frac{B}{x + 2}$

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

So $11 - 2 x = A \left(x + 2\right) + B \left(x - 1\right)$

If $x = 1$ $\implies$ $11 - 2 = 3 A + 0$
so $A = \frac{9}{3} = 3$

If $x = - 2$ $\implies$ $11 + 4 = 0 - 3 B$
so $B = \frac{15}{-} 3 = - 5$

$\frac{11 - 2 x}{\left(x - 1\right) \left(x + 2\right)} = \frac{3}{x - 1} - \frac{5}{x + 2}$

Then we can integrate

$\int \frac{\left(11 - 2 x\right) \mathrm{dx}}{\left(x - 1\right) \left(x + 2\right)} = \int \frac{3 \mathrm{dx}}{x - 1} - \int \frac{5 \mathrm{dx}}{x + 2}$

$= 3 \ln \left(x - 1\right) - 5 \ln \left(x + 2\right) + C$