## Calculus (3rd Edition)

$$\frac{-x}{\sqrt{x^{2}-1}}+\ln |x+\sqrt{x^{2}-1}|+C$$
Given $$\int \frac{x^{2}}{\left(x^{2}-1\right)^{3 / 2}} d x$$ Let $$x=\sec u\ \ \ \ \ \to\ \ \ dx=\sec u \tan udu$$ Then \begin{align*} \int \frac{x^{2}}{\left(x^{2}-1\right)^{3 / 2}} d x&=\int \frac{\sec ^{2} u(\sec u \tan u) d u}{\left(\sec ^{2} u-1\right)^{3 / 2}}\\ &= \int \frac{\sec ^{2} u(\sec u \tan u) d u}{\left(\tan ^{2} u\right)^{3 / 2}}\\ &=\int \frac{\sec ^{3} u d u}{ \tan ^{2} u }\\ &=\int \sec u\csc^2 udu \end{align*} Use integration by parts with $$U= \sec udu,\ \ \ dv=\csc^2 udu$$ Then \begin{align*} \int \frac{x^{2}}{\left(x^{2}-1\right)^{3 / 2}} d x&=\int \frac{\sec ^{2} u(\sec u \tan u) d u}{\left(\sec ^{2} u-1\right)^{3 / 2}}\\ &= \int \frac{\sec ^{2} u(\sec u \tan u) d u}{\left(\tan ^{2} u\right)^{3 / 2}}\\ &=\int \frac{\sec ^{3} u d u}{ \tan ^{2} u }\\ &=\int \sec u\csc^2 udu\\ &=-\sec u \cot u+\int \sec u \tan u \cot u \, d u\\ &=-\sec u \cot u+\int \sec u \, d u\\ &=-\sec u \cot u+\ln |\sec u+\tan u|+C\\ &= \frac{-x}{\sqrt{x^{2}-1}}+\ln |x+\sqrt{x^{2}-1}|+C \end{align*}