# How do you integrate int (5x - 4 ) / (x^2 -4x) dx using partial fractions?

Jan 3, 2016

Use partial fractions to give the expression in a form that you can integrate, and then integrate it normally.
Answer: $\ln x + 4 \ln \left(x - 4\right) + c$

#### Explanation:

Partial Fraction
Initially, ignore the fact that you will be integrating.
Let
$f \left(x\right) = \frac{5 x - 4}{{x}^{2} - 4 x}$

Factorise the denominator:

$f \left(x\right) = \frac{5 x - 4}{x \left(x - 4\right)}$

Then let

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

So

$\frac{5 x - 4}{x \left(x - 4\right)} = \frac{A}{x} + \frac{B}{x - 4}$

Then multiply by the denominator of $f \left(x\right)$

$5 x - 4 = \frac{x \left(x - 4\right) A}{x} + \frac{x \left(x - 4\right) B}{x - 4}$

This causes the cancellation of some terms:

$5 x - 4 = A \left(x - 4\right) + B x$

To find $A$ and $B$, you can use substitutions or compare coefficients. I prefer comparing coefficients for small problems, but using a mixture of the two techniques is often helpful.

Substitution
let $x = 4$
Therefore:
$20 - 4 = 0 + 4 B$
$16 = 4 B$
$B = 4$

Comparing Coefficients
For the constant terms:
$- 4 = - 4 A$
$A = 1$

Now we have everything we need to form our partial fraction:

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

$f \left(x\right) = \frac{1}{x} + \frac{4}{x - 4}$

We are now ready to integrate.

Integration

$\int \frac{5 x - 4}{{x}^{2} - 4 x} \mathrm{dx} = \int \frac{1}{x} + \frac{4}{x - 4} \mathrm{dx}$

It is easier to see if we separate the two parts of the integration:

$\int \frac{1}{x} \mathrm{dx} + \int \frac{4}{x - 4} \mathrm{dx}$

These both integrate in the same manner, using the standard antiderivative rule:

$\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$

$\ln x + 4 \ln \left(x - 4\right) + c$