inverse trigonometric integral

$\int \frac{1}{\sqrt{a^2+u^2}}du={\sin }^{-1}\frac{u}{a}+C$ $\int \frac{1}{a^2+u^2}du=\frac{1}{a}{\tan }^{-1}\frac{u}{a}+C$ $\int \frac{1}{u\sqrt{u^2-a^2}}du=\frac{1}{a}{\sec }^{-1}\left|\frac{u}{a}\right|+C$