How do you prove the integral formula intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C ?

1 Answer
Sep 11, 2014

Let x=atan theta. Rightarrow dx=asec^2theta d theta
So, we can write
int{dx}/{sqrt{x^2+a^2}}=int{asec^2theta}/{sqrt{a^2(tan^2theta+1)}}d theta
by tan^2theta+1=sec^2theta,
=intsec theta d theta=ln|sec theta+tan theta|+C
since tan theta=x/a and sec theta={sqrt{x^2+a^2}}/a,
=ln|{sqrt{x^2+a^2}+x}/a|+C_1
by the log property ln{A/B}=lnA-lnB,
=ln|sqrt{x^2+a^2}+x|-ln|a|+C_1
by setting C=ln|a|+C_1,
=ln|x+sqrt{x^2+a^2}|+C