# How do you evaluate the integral int dx/(3x(x-4))?

Mar 15, 2018

$\frac{1}{12} \ln | \frac{x - 4}{x} | + C , \mathmr{and} , \frac{1}{12} \ln | \left(1 - \frac{4}{x}\right) | + C .$

#### Explanation:

Let, $I = \int \frac{\mathrm{dx}}{3 x \left(x - 4\right)} = \frac{1}{3} \int \frac{1}{x \left(x - 4\right)} \mathrm{dx}$.

$\therefore I = \frac{1}{3} \cdot \frac{1}{4} \int \frac{4}{x \left(x - 4\right)} \mathrm{dx}$,

$= \frac{1}{12} \int \frac{x - \left(x - 4\right)}{x \left(x - 4\right)} \mathrm{dx}$,

$= \frac{1}{12} \int \left[\frac{x}{x \left(x - 4\right)} - \frac{x - 4}{x \left(x - 4\right)}\right] \mathrm{dx}$,

$= \frac{1}{12} \int \left[\frac{1}{x - 4} - \frac{1}{x}\right] \mathrm{dx}$,

$= \frac{1}{12} \left[\ln | \left(x - 4\right) | - \ln | x |\right]$.

$\Rightarrow I = \frac{1}{12} \ln | \frac{x - 4}{x} | + C , \mathmr{and} , \frac{1}{12} \ln | \left(1 - \frac{4}{x}\right) | + C .$

OTHERWISE,

If we use the formula $\int \frac{1}{{x}^{2} - {a}^{2}} \mathrm{dx} = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + c$,

then, $I = \frac{1}{3} \int \frac{1}{{x}^{2} - 4 x} \mathrm{dx} = \frac{1}{3} \int \frac{1}{{\left(x - 2\right)}^{2} - {2}^{2}} \mathrm{dx}$,

$= \frac{1}{3} \cdot \frac{1}{2 \cdot 2} \ln | \frac{\left(x - 2\right) - 2}{\left(x - 2\right) + 2} |$,

$= \frac{1}{12} \ln | \frac{x - 4}{x} | + C$, as before!