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