Question

Prove: $\int \frac{dx}{\sqrt{x^2-a^2}}=\log \left(\frac{x+\sqrt{x^2-a^2}}{a}\right)+c$

Answer 1

$\int \frac{1}{\sqrt{x^2-a^2}}dx$

substitute $\mu =\frac{x}{a}\to dx=adu$

$\Rightarrow \int \frac{a}{\sqrt{a^2{x}^2-a^2}}du$

Simplifying, take a common from denominator

$\Rightarrow \int \frac{1}{\sqrt{u^2-1}}du$

Subsitute $u=\sec \left(v\right),\left(v\right)=arc\sec \left(u\right)$

$du=\sec \left(v\right)\tan \left(v\right)dv$

$\Rightarrow \int \frac{\sec \left(v\right)\tan \left(v\right)dv}{\sqrt{{\sec }^2\left(v\right)-1}}$

Simply using ${\sec }^2\left(v\right)=1-{\tan }^2\left(v\right)$

$\Rightarrow \int \sec \left(v\right)dv$

Use the common Integral

$\Rightarrow \int \frac{\sec }{(}v)dv=\ln |\tan (v)+\sec (v)|$

Substitute back $v=arc\sec \left(u\right)$

$\Rightarrow \ln |\tan \left(arc\sec \right(u\left)\right)+\sec \left(arc\sec \right(u\left)\right)|$

using $\sec \left(arc\sec \right(u\left)\right)=u$

and using $\tan \left(arc\sec \right(u\left)\right)=u\sqrt{1-\frac{1}{u^2}}$

$\Rightarrow \ln |u+\mu \sqrt{1-\frac{1}{u^2}}|+C$

now put $u\Rightarrow \frac{x}{a}$

and we get

$\ln |\frac{\sqrt{x^2-a^2+x}}{a}|+C$