∫[(e^2x + 1)/(e^2x - 1)] dx =?

Question

possible solutions for homework problem...

a. (1/4)ln (1 + 2e2x) + C

b. -(1/4)ln |1 - 2e2x | + C

c. -(1/2)ln |2 - e^2x| + C

d. (1/2)ln (2 + e^2x) + C

e. ln| e^x – e^-x| + C

f. ln| e^x + e^-x| + C

g. x – e^-2x + C

h. x + e^2x + C

or is it none of these solutions?

Answer 1

Let u=e^(2x) - 1. Then du=2e^(2x) dx = 2(e^(2x) - 1) + 2 dx = 2u + 2 dx, so dx = 1/(2u+2) du

∫(u+2)/u * 1/(2u+2) du

This is now a rational function, so let's break it up:

∫u/(u(2u+2)) + 2/(u(2u+2)) du

Simplify a bit:

∫1/(2u+2) + 1/(u(u+1)) du

Rewrite the right fraction:

∫1/(2u+2) + (u+1-u)/(u(u+1)) du

Expand:

∫1/(2u+2) + (u+1)/(u(u+1)) - u/(u(u+1)) du

Simplify:

∫1/(2u+2) + 1/u - 1/(u+1) du

Simplify further:

∫1/u - (1/2)/(u+1) du

Now integrate:

ln |u| - 1/2 ln |u+1| + C

Resubstitute:

ln |e^(2x) - 1| - 1/2 ln |e^(2x)| + C

Simplify:

ln |e^(2x) - 1| - ln |e^x| + C

Simplify further:

ln |(e^(2x) - 1)/e^x| + C

And further:

ln |e^x - e^(-x)| + C

So the answer is e.

Answer 2

f