How do you find the antiderivative of #int 1/sqrt(1+x^2) dx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer mason m Jun 23, 2017 #lnabs(x+sqrt(1+x^2))+C# Explanation: #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# Answer link Related questions How do you find the integral #int1/(x^2*sqrt(x^2-9))dx# ? How do you find the integral #intx^3/(sqrt(x^2+9))dx# ? How do you find the integral #intx^3*sqrt(9-x^2)dx# ? How do you find the integral #intx^3/(sqrt(16-x^2))dx# ? How do you find the integral #intsqrt(x^2-1)/xdx# ? How do you find the integral #intsqrt(x^2-9)/x^3dx# ? How do you find the integral #intx/(sqrt(x^2+x+1))dx# ? How do you find the integral #intdt/(sqrt(t^2-6t+13))# ? How do you find the integral #intx*sqrt(1-x^4)dx# ? How do you prove the integral formula #intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C# ? See all questions in Integration by Trigonometric Substitution Impact of this question 262968 views around the world You can reuse this answer Creative Commons License