Evaluate  int \ 1/(x^2-2x) \ dx ?

Mar 17, 2017

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

Explanation:

We want to find:

$\int \setminus \frac{1}{{x}^{2} - 2 x} \setminus \mathrm{dx}$

If we examine the integrand we can decompose into partial fractions, as follows:

$\frac{1}{{x}^{2} - 2 x} = \frac{1}{x \left(x - 2\right)}$
$\text{ } = \frac{A}{x} + \frac{B}{x - 2}$
$\text{ } = \frac{A \left(x - 2\right) + B x}{x \left(x - 2\right)}$

And so:

$1 \equiv A \left(x - 2\right) + B x$

Put:

$x = 0 \implies 1 = - 2 A \implies A = - \frac{1}{2}$
$x = 2 \implies 1 = 2 B \setminus \setminus \setminus \setminus \setminus \implies B = \frac{1}{2}$

Therefore:

$\int \setminus \frac{1}{{x}^{2} - 2 x} \setminus \mathrm{dx} = \int \setminus \frac{\frac{1}{2}}{x - 2} - \frac{\frac{1}{2}}{x} \setminus \mathrm{dx}$
$\text{ } = \frac{1}{2} \ln | x - 2 | - \frac{1}{2} \ln | x | + C$
$\text{ } = \frac{1}{2} \ln | \frac{x - 2}{x} | + C$
$\text{ } = \frac{1}{2} \ln | \frac{A \left(x - 2\right)}{x} |$