Integrate: root(1 + e^x)dx?

Question

Please show work.
- Thanks

Answer 1

Let u=√(1+e^x), then du=e^x/(2√(1+e^x)) dx, so dx = 2√(1+e^x)/e^x du = 2u/(u²-1) du. Then we have:

∫u * 2u/(u²-1) du

Ah, much nicer, this is just a rational function. Simplifying a bit:

2∫u²/(u²-1) du
2∫(u²-1)/(u²-1) + 1/(u²-1) du
2∫1 + 1/(u²-1) du

We are almost ready to integrate, we just need the partial fraction decomposition of 1/(u²-1):

1/(u²-1) = A/(u-1) + B/(u+1)
1 = (u+1)A + (u-1)B

Substituting u=1 yields 2A=1, so A=1/2, Substituting u=-1 yields -2B=1, so B=-1/2. Thus we have:

2∫1 + (1/2)/(u-1) - (1/2)/(u+1) du
∫2 + 1/(u-1) - 1/(u+1) du
2u + ln |u-1| - ln |u+1| + C

Now just back-substitute, and we are done:

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