Calculus 10th Edition

$\frac{x(x-8)+ln|(x^2-5)|}{2}+C$
$\int \frac{x^3-4x^2-4x+20}{x^2-5}dx$ //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 $x^2-5=y$ $2x$ $dx=du$ $\int\frac{x}{x^2-5}dx$ $=\frac{1}{2}\int \frac{1}{u}du$ $=\frac{1}{2}ln|u|$ $=\frac{1}{2}ln|(x^2-5)|$ $=\int (x-4)dx$ $+\int\frac{x}{x^2-5}dx$ $=\frac{x^2}{2}-4x+\frac{ln|(x^2-5)|}{2}+C$ $=\frac{x(x-8)+ln|(x^2-5)|}{2}+C$