# How do you integrate int (2x - 1) / (x^2 + 5x - 6) dx using partial fractions?

Oct 31, 2016

The integral equals $\frac{13}{7} \ln | x + 6 | + \frac{1}{7} \ln | x - 1 | + C$.

#### Explanation:

We can factor the denominator as $\left(x + 6\right) \left(x - 1\right)$.

So,

$\frac{A}{x + 6} + \frac{B}{x - 1} = \frac{2 x - 1}{\left(x + 6\right) \left(x - 1\right)}$

$A \left(x - 1\right) + B \left(x + 6\right) = 2 x - 1$

$A x - A + B x + 6 B = 2 x - 1$

$\left(A + B\right) x + \left(- A + 6 B\right) = 2 x - 1$

We can now write a systems of equations to solve for $A$ and $B$.

$A + B = 2$
$- A + 6 B = - 1$

$B = 2 - a$

$- A + 6 \left(2 - A\right) = - 1$

$- A + 12 - 6 A = - 1$

$- 7 A = - 13$

$A = \frac{13}{7}$

$\frac{13}{7} + B = 2$

$B = 2 - \frac{13}{7}$

$B = \frac{1}{7}$

Resubstitute into the initial equation.

The integral become $\int \left(\frac{\frac{13}{7}}{x + 6} + \frac{\frac{1}{7}}{x - 1}\right) \mathrm{dx}$

We now use the rule $\int \left(\frac{1}{x}\right) \left(\mathrm{dx}\right) = \ln | x | + C$ to get rid of the integral.

Letting the function be $f \left(x\right)$.

$f \left(x\right) = \frac{13}{7} \ln | x + 6 | + \frac{1}{7} \ln | x - 1 | + C$

Hopefully this helps!