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 $sectheta=sqrt(1+tan^2theta)=sqrt(1+x^2)$ :

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

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