How do you integrate #int 1/sqrt(x^2+1)# by trigonometric substitution? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Ratnaker Mehta Aug 14, 2016 #ln|x+sqrt(x^2+1)|+C#. Explanation: We take subst. #x=tant rArr dx= sec^2tdt# Also, #1/sqrt(x^2+1)=1/sqrt(tan^t+1)=1/sect =cost# Hence, #I=int1/sqrt(x^2+1)dt# #=intcostsec^2tdt# #=intsectdt# #=ln|sect+tant|# #=ln|x+sqrt(x^2+1)|+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 1404 views around the world You can reuse this answer Creative Commons License