Let #x=a\sec\theta\implies dx=a\sec\theta\tan\theta\ d\theta#
#\therefore I=\int \sqrt{x^2-a^2}\ dx#
#I=\int \sqrt{a^2\sec^2\theta-a^2}\ a\sec\theta\tan\theta\ d\theta#
#I=\int (a\tan\theta)a\sec\theta\tan\theta\ d\theta#
#I=a^2\int \sec\theta\tan^2\theta\ d\theta\ .........(1)#
#I=a^2\int \tan\theta (\sec\theta\tan\theta)\ d\theta#
#I=a^2\tan\theta\int \sec\theta\tan\theta\ d\theta-a^2\int (d/{d\theta}\tan\theta\cdot \int \sec\theta\tan\theta\ d\theta)d\theta#
#I=a^2\tan\theta\sec\theta-a^2\int sec^2\theta \sec\theta\ d\theta#
#I=a^2\sec\theta\tan\theta-a^2\int (1+tan^2\theta) \sec\theta\ d\theta#
#I=a^2\sec\theta\tan\theta-a^2\int \sec\theta\ d\theta-a^2\int \sec\theta\tan^2\theta\ d\theta#
#I=a^2\sec\theta\tan\theta-a^2\ln|\sec\theta+\tan\theta|-I\ \quad (\text{put I from(1)})#
#2I=a^2\sec\theta\tan\theta-a^2\ln|\sec\theta+\tan\theta|+c#
#I=1/2(a^2\sec\theta\tan\theta-a^2\ln|\sec\theta+\tan\theta|)+c_1#
#I=1/2(a^2(x/a)\sqrt{x^2/a^2-1}-a^2\ln|x/a+\sqrt{x^2/a^2-1}|)+c_1#
#I=1/2(x\sqrt{x^2-a^2}-a^2\ln|x+\sqrt{x^2-a^2}|+a^2\ln a)+c_1#
#I=x/2 \sqrt{x^2-a^2}-{a^2}/2\ln|x+\sqrt{x^2-a^2}|+C#
Proved.
