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?
Let $x$ be positive. Then $$\int_1^x \frac{1}{t}\,dt=\ln x.$$ Let us try to relate this to the fact that $$\int_1^x t^r\,dt=\frac{1}{r+1}\left(x^{r+1}-1\right)$$ if $r\ne -1$. We will calculate $$\lim_{r\to -1} \frac{x^{r+1}-1}{r+1}.$$ There are many ways to calculate the limit. Let us use the fact that $x^{r+1}=e^{(r+1)\ln x}$. So we want $$\lim_{r\to -1} \frac{e^{(r+1)\ln x}-1}{r+1}.$$ Using say L'Hospital's Rule, we find that this limit is equal to $$\lim_{r\to -1} \frac{(\ln x)e^{(r+1)\ln x}}{1},$$ which is $\ln x$.
$\int x^{n} dx = \frac{x^n+1}{n+1}$ is true only for $n\neq-1$ For $n=-1, \int x^{n} dx = \int\frac{1}{x} dx = \ln(x) + c$
The "rule" $\int x^\alpha \ \mathrm{d}x = \frac{x^{\alpha+1}}{\alpha+1}$ is good only for $\alpha \neq -1$. For $\alpha = -1$, we have: $$\int \frac{1}{x} \ \mathrm{d}x = \ln x + c, \qquad c \in \Bbb R.$$ Indeed, $\frac{\mathrm{d}}{\mathrm{d}x}\ln x = \frac{1}{x}$.
$\int \frac{1}{x}dx$ is defined to be $ln(|x|)$: http://www.math-prof.com/Calculus_1/Calc_Ch_26.asp
