Evaluate the Integral: $\int_0^2 \frac{dx}{e^{\pi x}}$ 
Evaluate the definite integral: $$\int_0^2 \frac{\mathrm{d}x}{e^{\pi x}}$$

My attempt:
$u=e^{\pi x}$
$du=\pi e^{\pi x}\ dx$
$\int_0^2 \frac{1}{u}\ dx$ 
So at this point do I divide $\pi e^{\pi x}$ by dx?
Thus, $\frac{du}{\pi\ e^{\pi x}}=dx$
and $\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}}$
and $\frac{1}{u}\int_0^2\frac{du}{\pi\ e^{\pi x}} $ 
Find the new values for integral? I have no idea how to do this. 
I basically would like to know if my process of solving this problem is correct and how to complete solving the problem.  

note that in $\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}}$ we can substitute $\pi\ e^{\pi x}$ with $\pi u$. Thus $$\int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ e^{\pi x}} = \int_0^2 \frac{1}{u}\cdot \frac{du}{\pi\ u} = \int_0^2 \frac{du}{\pi u^2} = \frac{1}{\pi}\left[ -\frac{1}{u} \right]^{u=2}_{u=0}$$.
However you forgot to change the range of the integral with the substitution with $u=e^{\pi x}$ thus this does not work. $x=0 \rightarrow u= 1$ and $x=2 \rightarrow u= e^{2 \pi}$. You have to put these in so we get:
$$\frac{1}{\pi}\left[ -\frac{1}{u} \right]^{u=e^{2 \pi}}_{u=1} = \frac{-1}{\pi e^{2 \pi}} +\frac{1}{\pi}  $$
