Question

Find the integral of $\frac{1}{\sqrt{x^2+a^2}}$ with respect to x

Answer 1

$I=\int \frac{dx}{\sqrt{x^2+a^2}}$

put $x=a\tan \theta ⟹dx=a{\text{sec}}^2\theta d\theta$

$I=\int \frac{a{\text{sec}}^2\theta d\theta }{\sqrt{a^2+a^2{\tan }^2\theta }}=\int \text{sec}\theta d\theta =\ln |\text{sec}\theta +\tan \theta |+C=\ln |\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}}|+C=\ln |x+\sqrt{x^2+a^2}|+C$