How do you integrate (x+2)/(2x^3-8x) using partial fractions?

Oct 30, 2017

$\int \setminus \frac{x + 2}{2 {x}^{3} - 8 x} \setminus \mathrm{dx} = \frac{1}{4} \ln | \frac{x - 2}{x} | + C$

Explanation:

We seek:

$I = \int \setminus \frac{x + 2}{2 {x}^{3} - 8 x} \setminus \mathrm{dx}$

We can write as:

$I = \int \setminus \frac{x + 2}{2 x \left({x}^{2} - 4\right)} \setminus \mathrm{dx}$
 \ \ = 1/2 \ int \ (x+2)/(x(x+2)(x-2) \ dx (removable discontinuity)
 \ \ = 1/2 \ int \ 1/(x(x-2) \ dx

We can now decompose the integrand in to partial fractions:

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

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

Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

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

So we can now write:

$I = \frac{1}{2} \setminus \int \setminus \frac{- \frac{1}{2}}{x} + \frac{\frac{1}{2}}{x - 2} \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{4} \setminus \int \setminus \frac{1}{x - 2} - \frac{1}{x} \setminus \mathrm{dx}$

Which now consists of standard integral so we integrate to get:

$I = \frac{1}{4} \left\{\ln | x - 2 | - \ln | x |\right\} + C$
$\setminus \setminus = \frac{1}{4} \ln | \frac{x - 2}{x} | + C$