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

Mar 22, 2018

The integral equals $3 \ln | x + 2 | - 2 \ln | x + 1 | + C$

#### Explanation:

Note that the factoring of ${x}^{2} + 3 x + 2$ is $\left(x + 2\right) \left(x + 1\right)$. Therefore:

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

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

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

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

We now have a system of equations: $\left\{\begin{matrix}A + B = 1 \\ A + 2 B = - 1\end{matrix}\right.$

We can readily solve through elimination (subtract the second equation from the first to get the following):

$- B = 2$

$B = - 2$

It is now clear that $A - 2 = 1 \to A = 3$. The integral becomes

$I = \int \frac{3}{x + 2} - \frac{2}{x + 1} \mathrm{dx}$

We can now easily integrate.

$I = 3 \ln | x + 2 | - 2 \ln | x + 1 | + C$

Hopefully this helps!