How do you find the antiderivative of #e^(2x) * sqrt(e^x + 1)#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Ratnaker Mehta Sep 18, 2016 #2/15(e^x+1)^(3/2)(3e^x-2)+C#. Explanation: Let, #I=inte^(2x)sqrt(e^x+1)dx# We use the subst. #e^x+1=t^2, or, e^x=t^2-1", so that, "e^xdx=2tdt#. #:. I=inte^xsqrt(e^x+1)*e^xdx# #=int(t^2-1)sqrt(t^2)*2tdt# #=2int(t^4-t^2)dt# #=2(t^5/5-t^3/3)# #=2/15(3t^5-5t^3)# #=2/15t^3(3t^2-5)# #=2/15(e^x+1)^(3/2)({3(e^x+1)-5}..............[as, t=(e^x+1)^(1/2)]# #=2/15(e^x+1)^(3/2)(3e^x-2)+C#. Answer link Related questions How do you evaluate the integral #inte^(4x) dx#? How do you evaluate the integral #inte^(-x) dx#? How do you evaluate the integral #int3^(x) dx#? How do you evaluate the integral #int3e^(x)-5e^(2x) dx#? How do you evaluate the integral #int10^(-x) dx#? What is the integral of #e^(x^3)#? What is the integral of #e^(0.5x)#? What is the integral of #e^(2x)#? What is the integral of #e^(7x)#? What is the integral of #2e^(2x)#? See all questions in Integrals of Exponential Functions Impact of this question 4681 views around the world You can reuse this answer Creative Commons License