Question: I was thinking about $\int \frac{1}{x}dx$ and how it is defined because if $\int x^{n} dx = \frac{x^n+1}{n+1}$ where $n$ is a constant then:
$$\int \frac{1}{x}dx = \frac{x^{-1+1}}{-1+1}$$
$$ = \frac{x^{0}}{0} $$
$$= \frac{1}{0}$$ since $x^{0} = 1$ and $\frac{1}{0}$ is not defined.

My question is how is it that $\int \frac{1}{x}dx$ exists if this is the result that I get?

Answer: 
$\int x^{n} dx = \frac{x^n+1}{n+1}$ is true only for $n\neq-1$
For $n=-1, \int x^{n} dx = \int\frac{1}{x} dx = \ln(x) + c$
$\int \frac{1}{x}dx$ is defined to be $ln(|x|)$: <http://www.math-prof.com/Calculus_1/Calc_Ch_26.asp>