How do you find the antiderivative of #e^(2x) * sqrt(e^x + 1)#?

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Answer 1 · 2016-09-18T20:47:51

#2/15(e^x+1)^(3/2)(3e^x-2)+C#.

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#.