## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{2} x^2 + \dfrac{1}{2} \ln|x^2-1|+C$
Since, we have $\int \dfrac{x^3}{x^2-1} dx$ Now, the given integral can be written as: $\int \dfrac{x^3}{x^2-1} dx=\int x dx +\int \dfrac{x}{x^2-1} dx$ Consider $u=x^2-1 \implies du=2x dx$ for the second integral and integrate. $\int x dx +\int \dfrac{x}{x^2-1} dx= \dfrac{1}{2}[ x^2+\int (1/u) du]+C$ or, $=\dfrac{1}{2} [x^2+\ln |u|]+C$ Again back plug in $u=x^2-1$: Hence, $= \dfrac{1}{2} x^2 + \dfrac{1}{2} \ln|x^2-1|+C$