How do you evaluate the integral #int e^(2x)sqrt(1+e^(2x))#?

1 Answer
Mar 1, 2017

Answer:

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

Explanation:

Let #u = 1 + e^(2x)#. Then #du = 2e^(2x)dx# and #dx = (du)/(2e^(2x))#.

#int e^(2x)sqrt(u) * (du)/(2e^(2x))#

#1/2int sqrt(u) du#

#1/2(2/3u^(3/2)) + C#

#1/3u^(3/2) + C#

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

Hopefully this helps!