How do you find #\int e ^ { 2x } \sqrt { 1+ e ^ { 2x } } d x#?

1 Answer
NickTheTurtle
Apr 13, 2017

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

Explanation:

The question asks for #int\ e^(2x)sqrt(1+e^(2x))\ dx#.

Substitute #u=1+e^(2x)#. Then, #(du)/dx=2e^(2x)#, or (solving for #dx#) #dx=(du)/(2e^(2x))#.

Then, #int\ e^(2x)sqrt(1+e^(2x))\ dx=int\ e^(2x)sqrt(u)\ (du)/(2e^(2x))#.

Simplify to get #1/2int\ sqrt(u)\ du=1/2int\ u^(1/2)\ du#. This is a simple integral. Use the power rule #int\ x^n\ dx=x^(n+1)/(n+1)+C#.

Then, #1/2int\ u^(1/2)\ du=(u^(3/2))/3+C=(sqrt(u^3))/3+C#. We want our answer to be in terms of #x#. Previously, #u# was defined as #1+e^(2x)#. Just substitute this in to get #(sqrt((e^(2x)+1)^3))/3+C#.

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