# How do you find int (x - 4 ) / (x^2 -4x) dx using partial fractions?

Nov 4, 2015

$\int \frac{x - 4}{{x}^{2} - 4 x} \mathrm{dx} = \ln | x |$

#### Explanation:

$\int \frac{x - 4}{{x}^{2} - 4 x} \mathrm{dx} = \int \frac{x - 4}{x \left(x - 4\right)} \mathrm{dx} = \int \frac{1}{x} \mathrm{dx} = \ln | x |$

Nov 4, 2015

Reduce the fraction.

#### Explanation:

When you factor the denominator to start the partial fraction decomposition, note that the ratio can be reduced.

$\int \frac{x - 4}{{x}^{2} - 4 x} \mathrm{dx} = \int \frac{x - 4}{x \left(x - 4\right)} \mathrm{dx}$

$= \int \frac{1}{x} \mathrm{dx} = \ln \left\mid x \right\mid + C$

If you didn't notice it could be reduced, find $A , \text{and } B$ so that:

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

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

And $A + B = 1$

and $- 4 A = - 4$, $\text{ }$ so $A = 1$ and $B = 0$