Evaluating $\;\int_{1}^{\ln3}\frac{e^x - e^{2x}}{(1 + e^x)}\,dx$ 
Find $\int_{1}^{\ln3}(e^x - e^{2x})/(1 + e^x)dx$.

I looked through my notes for integration techniques and thought I could try a $u$ substitution but whatever I set $u$ to I can't seem to simplify this integral. Any hints on how I could solve this?
 A: Let $u = 1+ e^x$, and $e^x = u - 1$.  
Then $du = e^x\,dx$. Note that your numerator factors into $$(e^x - (e^x)^2)\,dx = e^x(1-e^x)\,dx = (1- (u -1)) \underbrace{e^x\,dx}_{\large du} = (2-u)\,du$$
New limits of integration: At $\,x=1, \;u = 1+e\,;\quad$ at $\,x = \ln 3,\; u = 1+ 3 = 4$.
That gives you the integral $$\int_{1+e}^4\frac{(2-u)\,du}{u} =  \int_{1+e}^4 \frac 2u\,du -\int_{1+e}^4  \,du $$
with no need to back-substitute, since we adjusted the limits of integration.
A: HINT : Setting $u=1+e^x$ gives you $e^xdx=du$ and,$$\int\frac{e^x-e^{2x}}{1+e^x}dx=\int\frac{(1-e^x)\color{red}{e^xdx}}{1+e^x}=\int\frac{1-(u-1)}{u}\color{red}{du}=\int\left(-1+\frac 2u\right)du.$$
A: $$I=\int_1^{\log3}\frac{e^x-e^{2x}}{1+e^x}dx$$
$$I=\int_1^{\log3}\frac{e^x(1-e^x)}{1+e^x}dx$$
$u=e^x\Rightarrow du=e^xdx$
$$I=\int_e^3\frac{1-u}{1+u}du$$
$$I=\int_e^3\frac{du}{1+u}-\int_e^3\frac{u}{1+u}du$$
$$I=\log(1+u)\Bigg|_e^3-\int_e^3\frac{u}{1+u}du$$
$$I=\log4-\log(1+e)-\int_e^3\frac{u}{1+u}du$$
$w=u+1\Rightarrow dw=du$
$$I=\log\frac4{1+e}-\int_{1+e}^4 \frac{w-1}w dw$$
$$I=\log\frac4{1+e}+e-3+\log w\bigg|_{1+e}^4$$
$$I=2\log\frac4{1+e}+e-3$$
