Integral of 1/x- why does it behave this way? First of all, I'm not entirely sure what to call polynomial functions that can have x raised to negative integers powers, so I will just call them polynomial type functions.
The integral of every polynomial type function is another polynomial type function, unless, of course, our polynomial type function has $\frac{1}{x}$ in it. In that case, our integral is obviously $\ln{x}$.
In the case of every other function in the form $x^{k}$ for some integer $k$, we can use power rule to find the integral. With $\frac{1}{x}$, there is a problem, since our integral will have had a constant of $0$ multiplying the term, nullifying it.
So my question is, why does it happen to be true that every single polynomial type function has a polynomial type integral, except $\frac{1}{x}$? $\frac{1}{x}$ happens to be the only function where using power rule to evaluate the integral doesn't work, and it also happens to the be the only one without a polynomial type integral.
This can't be a coincidence: the two special properties of $\frac{1}{x}$ must be related. What about $\frac{1}{x}$ makes it so special?
 A: One way to see it is that the derivative of $\ln(x)$ is equal to $\frac{1}{x}$:
$$y=\ln(x)$$ $$e^y=x$$ Differentiating both sides with respect to x,
$$\frac{dy}{dx}e^y=1$$ $$\frac{dy}{dx}=\frac{1}{e^y}$$
But remember that $e^y=x$. This means that $$\frac{dy}{dx}=\frac{1}{x}$$
This means that the integral of $\frac{1}{x}$ has to be $\ln(x)$.
I'm not sure what you mean when you say that $\frac{1}{x}$ doesn't have a "polynomial type" derivative. From your definition $\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$ is a polynomial type function.
A: $\frac 1x$ does have a polynomial type derivative.  The derivative is $-\frac 1{x^2}$ which is polynomial type.  This is a central assumption of the question which is wrong.
A: The integral of $\frac{1}{x}$ is not a rational function. Several proofs of this can be found on this website. There are however power series expansions of $\ln(x)$ for some but not all of the real line. Recall the geometric series:
\begin{align*}
1+x+x^2+x^3+\cdots=\frac{1}{1-x}
\end{align*}
This series converges for $|x|<1$. Integrating both sides yields
\begin{align*}
\int(1+x+x^2+x^3+\cdots)\,dx&=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots\\
&=\sum_{k=1}^\infty \frac{x^k}{k}.
\end{align*}
Therefore
\begin{align*}
\int\frac{1}{1-x}\,dx=\sum_{k=1}^\infty \frac{x^k}{k}
\end{align*}
This is a but unwieldy though because of the negative sign, so flipping this around gives
\begin{align*}
\int\frac{1}{1+x}\,dx=\sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k}
\end{align*}
Using a change of variables and the definition of the natural logarithm, the power series on the RHS is equal to $\ln(1+x)$. So to summarize, the integral of $\frac{1}{x}$ can represented as a power series, but not a rational function. If you want to see some more cool stuff like this, you can follow this link.
