How do you find the integral of #e^(2x) sqrt(1 + e^(2x)) dx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer bp Jun 9, 2015 #1/3 (1+e^(2x))^(3/2)# +C Explanation: Let u=# 1+e^(2x)#, du= 2#e^(2x)#dx Accordingly, #int e^(2x) sqrt(1+e^(2x) dx#=# 1/2int u^(1/2) du# =#1/3 u^(3/2)# +C =#1/3 (1+e^(2x))^(3/2)# +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 4899 views around the world You can reuse this answer Creative Commons License