How do you find the antiderivative of #dx/(e^(2x)-9)^(1/2)#?

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Answer 1 · 2015-02-19T17:54:35

The answer is: #1/3arctan(sqrt(e^(2x)-9)/3)+c#.

We have to make this substitution:

#sqrt(e^(2x)-9)=trArre^(2x)-9=t^2rArre^(2x)=t^2+9rArr#

#2x=ln(t^2+9)rArrx=1/2ln(t^2+9)rArrdx=1/2*(2t)/(t^2+9)dtrArr#

#dx=t/(t^2+9)dt#,

so:

#int(dx)/sqrt(e^(2x)-9)=int1/t*t/(t^2+9)dt=intdt/(t^2+9)=#

#intdt/(9(t^2/9+1))=1/9intdt/((t/3)^2+1)=1/9*3int(1/3)/((t/3)^2+1)dt=#

#=1/3arctan(t/3)+c=1/3arctan(sqrt(e^(2x)-9)/3)+c#.