Question

Find the integral of $\frac{1}{\sqrt{x^2-a^2}}$ with respect to $x$ and hence evaluate
$\int \frac{dx}{\sqrt{x^2+6x-7}}$.

Answer 1

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

$x=a\sec \theta ,dx=a\sec \theta \tan \theta d\theta$

$\sqrt{x^2-a^2}=a\tan \theta$

$I=\int \frac{a\sec \theta \tan \theta d\theta }{a\tan \theta }$

$I=\int \tan \theta d\theta =\ln |\sec \theta |+c$

$I=\int \frac{a\sec \theta \tan \theta d\theta }{a\tan \theta }$

$I=\int \sec \theta d\theta =\ln |\sec \theta +\tan \theta |+c$

$I=\ln (\frac{x}{a}+\sqrt{{\left(\frac{x}{a}\right)}^2-1})+c$

$I_1=\int \frac{dx}{\sqrt{x^2+6x-7}}=\int \frac{dx}{\sqrt{{(x+3)}^2-4^2}}$

$x=x+3,a=4$

$I=\ln (\frac{x+3}{4}+\sqrt{{\left(\frac{x+3}{4}\right)}^2-1})+c$