Answer
$\frac{x(x-8)+ln|(x^2-5)|}{2}+C$
Work Step by Step
$\int \frac{x^3-4x^2-4x+20}{x^2-5}dx$
//divide the numerator with the denominator
$=\int (x-4)dx$ $+\int\frac{x}{x^2-5}dx$
for $\int\frac{x}{x^2-5}dx$,
let $x^2-5=y$
$2x$ $dx=du$
$\int\frac{x}{x^2-5}dx$
$=\frac{1}{2}\int \frac{1}{u}du$
$=\frac{1}{2}ln|u|$
$=\frac{1}{2}ln|(x^2-5)|$
$=\int (x-4)dx$ $+\int\frac{x}{x^2-5}dx$
$=\frac{x^2}{2}-4x+\frac{ln|(x^2-5)|}{2}+C$
$=\frac{x(x-8)+ln|(x^2-5)|}{2}+C$
