How do you find the antiderivative of #int 1/sqrt(1+x^2) dx#?

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Answer 1 · 2017-06-23T04:01:37

#lnabs(x+sqrt(1+x^2))+C#

#I=int1/sqrt(1+x^2)dx#

Let #x=tantheta#. This implies that #dx=sec^2thetad theta#.

#I=int1/sqrt(1+tan^2theta)sec^2thetad theta#

Since #1+tan^2theta=sec^2theta#:

#I=intsecthetad theta=lnabs(sectheta+tantheta)#

Note that #tantheta=x# and #sectheta=sqrt(1+tan^2theta)=sqrt(1+x^2)#:

#I=lnabs(x+sqrt(1+x^2))+C#