How do you find#int x/sqrt(x^2 + 1) dx # using trigonometric substitution? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Cem Sentin Mar 12, 2018 #int x/sqrt(x^2+1)*dx=sqrt(x^2+1)+C# Explanation: #int x/sqrt(x^2+1)*dx# After using #x=tany# and #dx=(secy)^2*dy# transforms, this integral became #int tany/sqrt((tany)^2+1)*(secy)^2*dy# =#int tany/sqrt((secy)^2)*(secy)^2*dy# =#int ((secy)^2*tany)/secy*dy# =#int secy*tany*dy# =#secy+C# =#sqrt((tany)^2+1)+C# =#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 1360 views around the world You can reuse this answer Creative Commons License