How do you integrate #int e^x/sqrt(e^(2x)-4e^x+6)dx# using trigonometric substitution?

Question

4211 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T12:16:13

Use the substitution #e^x-2=sqrt2tantheta#.

Let

#I=inte^x/sqrt(e^(2x)-4e^x+6)dx#

Complete the square in the square root:

#I=inte^x/sqrt((e^x-2)^2+2)dx#

Apply the substitution #e^x-2=sqrt2tantheta#:

#I=intsecthetad theta#

Integrate directly:

#I=ln|sectheta+tantheta|+C#

Reverse the substitution:

#I=ln|(e^x-2)+sqrt((e^x-2)^2+2)|+C#