Question

Integrate $\int \sqrt{1+x^2}dx$ is equal to:

Answer 1

$={\int }^{{\sec }^2}\left(u\right)\sqrt{{\tan }^2\left(u\right)+1}du$

$={\int }^{{\sec }^3}\left(u\right)du$

$=\frac{\sec \left(u\right)\tan \left(u\right)}{2}+\frac{1}{2}\int sec\left(u\right)du$

$\int \sec \left(u\right)du$

$=\ln \left(\tan \left(u\right)+\sec \left(u\right)\right)$

$\frac{\sec \left(u\right)\tan \left(u\right)}{2}+\frac{1}{2}\int \sec \left(u\right)du$

$=\frac{\ln \left(\tan \left(u\right)+\sec \left(u\right)\right)}{2}+\frac{\sec \left(u\right)\tan \left(u\right)}{2}$

${\int }^{\sqrt{x^2+1}}dx$

$=\frac{\ln \left(\left|\sqrt{x^2+1}+x\right|\right)}{2}+\frac{x\sqrt{x^2+1}}{2}+C$

$\frac{ar\sinh \left(x\right)}{2}+\frac{x\sqrt{x^2+1}}{2}+C$