# How do you integrate int (4x-1)/( x^2(x-4)) using partial fractions?

Aug 3, 2016

$\int \frac{4 x - 1}{{x}^{2} \left(x - 4\right)} \mathrm{dx} = - \frac{15}{16} \ln x - \frac{1}{4 x} + \frac{15}{16} \ln \left(x - 4\right)$

#### Explanation:

Partial fractions of $\frac{4 x - 1}{{x}^{2} \left(x - 4\right)}$ will be of type $\frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x - 4}$ i.e.

$\frac{4 x - 1}{{x}^{2} \left(x - 4\right)} \Leftrightarrow \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x - 4}$ or

$\Leftrightarrow \frac{A x \left(x - 4\right) + B \left(x - 4\right) + C {x}^{2}}{{x}^{2} \left(x - 4\right)}$ or

$\Leftrightarrow \frac{A {x}^{2} - 4 A x + B x - 4 B + C {x}^{2}}{{x}^{2} \left(x - 4\right)}$ or

$\frac{4 x - 1}{{x}^{2} \left(x - 4\right)} \Leftrightarrow \frac{\left(A + C\right) {x}^{2} + x \left(- 4 A + B\right) - 4 B}{{x}^{2} \left(x - 4\right)}$

Hence $A + C = 0$, $- 4 A + B = 4$ and $4 B = 1$

i.e. $B = \frac{1}{4}$, $- 4 A = 4 - \frac{1}{4} = \frac{15}{4}$ or $A = - \frac{15}{16}$ and $C = \frac{15}{16}$ and

$\frac{4 x - 1}{{x}^{2} \left(x - 4\right)} \Leftrightarrow - \frac{15}{16 x} + \frac{1}{4 {x}^{2}} + \frac{15}{16 \left(x - 4\right)}$

Hence $\int \frac{4 x - 1}{{x}^{2} \left(x - 4\right)} \mathrm{dx} \Leftrightarrow \int - \frac{15}{16 x} \mathrm{dx} + \int \frac{1}{4 {x}^{2}} \mathrm{dx} + \int \frac{15}{16 \left(x - 4\right)} \mathrm{dx}$

= $- \frac{15}{16} \ln x - \frac{1}{4 x} + \frac{15}{16} \ln \left(x - 4\right)$