I need to show that this sequence is increasing and I'm almost there but I need help on last step. Let $(1+\frac{1}{n})^n$ be a sequence and $f(x)=(1+\frac{1}{x})^x $ on $[1,inf)$.  I need to show that f is non-decreasing by showing that $f'(x)\ge0$.  So far I have:

Let $g(x)=ln(f(x))$, where $ln$ means the natural log.  So $g'(x)=ln(1+\frac{1}{x})-\frac{1}{x+1}$.  Thus $g'(x)\ge0$ iff $ln(1+\frac{1}{x})\ge\frac{1}{x+1}$.  Which is equivalent to showing $\int_1 ^{1+1/x}\frac{1}{t}dt\ge\frac{1}{x+1}$.  Then I said let P be the trivial partition $({1,1+\frac{1}{x}})$.  So by definition of Riemann integrate, $L(P,1/t)=\frac{1}{x+1}$. 

This is where I get stuck.  What would be my next step.  

I am not looking for a different way to show this, this is how my teacher wants us to do this problem.
The interval $I=[1,1+1/x]$ has length $1/x$. At its lowest (on $I$), the function $t\mapsto 1/t$ takes on the value $\frac{1}{1+1/x}=\frac{x}{x+1}$, hence
$$\int_{1}^{1+1/x}\frac{dt}{t}\geq \frac{1}{x}\times\frac{x}{x+1}=\frac{1}{x+1}$$
The interval $I=[1,1+1/x]$ has length $1/x$. At its lowest (on $I$), the function $t\mapsto 1/t$ takes on the value $\frac{1}{1+1/x}=\frac{x}{x+1}$, hence
$$\int_{1}^{1+1/x}\frac{dt}{t}\geq \frac{1}{x}\times\frac{x}{x+1}=\frac{1}{x+1}$$