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

1 Answer
Wataru
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#

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