How do you use partial fractions to find the integral int (x+2)/(x^2-4x)dx?

Dec 28, 2016

First, factor the denominator.

${x}^{2} - 4 x = x \left(x - 4\right)$

$\frac{A}{x} + \frac{B}{x - 4} = \frac{x + 2}{\left(x\right) \left(x - 4\right)}$

$A \left(x - 4\right) + B \left(x\right) = x + 2$

$A x + B x - 4 A = x + 2$

$\left(A + B\right) x - 4 A = x + 2$

We can now write a systems of equations.

$\left\{\begin{matrix}A + B = 1 \\ - 4 A = 2\end{matrix}\right.$

Solving, we get $A = - \frac{1}{2}$ and $B = \frac{3}{2}$.

$\therefore$ The partial fraction decomposition is $\frac{3}{2 \left(x - 4\right)} - \frac{1}{2 x}$.

This can be integrated using the rule $\int \frac{1}{u} \mathrm{du} = \ln | u | + C$.

$= \frac{3}{2} \ln | x - 4 | - \frac{1}{2} \ln | x | + C$

Hopefully this helps!