Explain PLease

[Mr.Pacman]

OK i understand that


$\displaystyle \int\ e^3 / 1 +e^x dx = ln(1+e^x)$


but i dont understand why

if u let u = 1 + e^x
then du = e^x

then i get lost after this step
can u help me understand please

 

[galactus]

The e^3 is a constant, so bring it outside the integral sign.

Remember proper grouping symbols. You're in calc now.

$\displaystyle e^{3}\int\frac{1}{1+e^{x}}dx$

Let, as you correctly stated, $\displaystyle u=e^{x}, \;\ x=ln(u), \;\ dx=\frac{1}{u}du$

Then, we get:

$\displaystyle e^{3}\int\frac{1}{u+u^{2}}du$

Expand:

$\displaystyle e^{3}\left[\int\frac{1}{u}-\int\frac{1}{1+u}\right]du$

Intergate and resub

$\displaystyle e^{3}\left[ln(u)-ln(1+u)\right]$

$\displaystyle e^{3}\left[ln(e^{x})-ln(1+e^{x})\right]$

$\displaystyle e^{3}\left(x-ln(e^{x}+1)\right)$