How do you integrate #int e^(2x)/sqrt(e^(2x) +25)dx# using trigonometric substitution?

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Answer 1 · 2018-06-18T08:17:03

The answer is #=sqrt(e^(2x)+25)+C#

No need for trigonometric substitution.

Let #u=e^(2x)+25#, #=>#, #du=2e^(2x)dx#

Therefore, the integral is

#int(e^(2x)dx)/sqrt(e^(2x)+25)=1/2int(du)/(sqrt(u))#

#=1/2intu^(-1/2)du#

#=1/2u^(1/2)/(1/2)#

#=sqrtu#

#=sqrt(e^(2x)+25)+C#