I need help with integrating. [integral] sqrt (1+e^x)dx?

Question

I thought I knew how to do it..but I guess not. I tried using u = 1+e^x but I'm sooo confused.

Answer 1

The best way to approach this is probably to make u=√(1+e^x) dx. This gives you du=1/(2√(1+e^x)) e^x dx, which you can rewrite as du=1/(2u) (u²-1) dx, thus dx=2u/(u²-1) du. So your integral becomes:

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

You may then decompose 2/(u²-1) using partial fractions:

2/(u²-1)=A/(u+1)+B/(u-1)
2=A(u-1)+B(u+1)
2=(A+B)u + (B-A)
A+B=0, B-A=2
B=1, A=-1

So your integral becomes:

∫2 du + ∫1/(u-1) - 1/(u+1) du
2u + ln (u-1) - ln (u+1) + C
2√(1+e^x) + ln (√(1+e^x)-1)- ln (√(1+e^x)+1) + C

If you want to simplify this further, you can write:

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

Answer 2

let u = 1+ e^x
differentiate both sides
du = e^xdx= (u-1) dx
dx = du/(u-1)
integrtate
u^(1/2)du/(u-1) .... 1
let u = t^2
du = 2tdt
dt = du/2u^(1/2)= u^(1/2) du/2u = u^(1/2)du/2t^2
so
2t^2 dt = u^(1/2) du
so from 1

integrate 2t^2dt/(t^2-1)
= 2dt + 2/(t2^-1) = 2dt + dt/(t+1) - dt/(t-1)

intgral = 2t + ln(t+1) - ln(t-1)
you can now substitute and get in terms of x
I hope that I have not confused you more

=

Answer 3

∫(1 + e^x)^1/2
= (2/3(1+e^x) ^3/2) / e^x
= (2/3e^x) (√(1+e^x)³)

Another solution is
2√1 + e^x + 2 tanhˉ¹ (√1+e^x)

Answer 4

well dear ;

∫√(1 +e^x) dx = 2(√1 + e^x) -2 tanh^-1(√1+e^x) + c

{ tanh^-1 = arc tanh}
Good Luck.