Integral of monomial and logarithm: is this true? $\lim_{k\to -1}\frac{x^{k+1}}{k+1} = \log|x|$ It is well know that:
$$\int x^k \text{d}x = \begin{cases}
\displaystyle\frac{x^{k+1}}{k+1} + c & k \neq -1\\
\\
\log|x| + c & k = -1\end{cases}$$
My guess is:
$$\lim_{k\to -1}\frac{x^{k+1}}{k+1} = \log|x| ???$$
Apparently, this limit goes to infinity when $x>0$.
Having said that, is there something that "join" monomial to logarithm? I mean, why the integral of a monomial is a monomial except for the case $k=-1$?
Addition
I know very well that this is because 
$$\frac{\text{d}}{\text{d}x} \log(x) = \lim_{h \to 0^+} \frac{\log(x+h)-\log(x)}{h} =  \frac{1}{x}.$$
Anyway, the scheme "integral of monomial is a monomial" is somehow broken. What is the "deep" reason for this situation?
 A: your answer to the question is basically correct, so there is no convergence to $\log |x|$.
strictly speaking this limit does not go to infinity, because the one sided limits are $+\infty$ and $-\infty$. 
In order to derive a convergence of functions that are integrations of other functions $f_n$, you need uniform convergence, not pointwise convergence.
Edit: just an idea for your additional question: a monomial is zero at either $0$ or $\infty$, and a=-1 is the only value where both $\int_1^\infty x^a$ and $\int_0^1 x^a$ are infinite, therefor it is impossible that the integral of $x^{-1}$ is another monomial.
A: I must say its a very out of the box question and OP deserves credit to think in this way. +1 from my end. You are right but you need to express your ideas in concrete manner. You know that indefinite integrals are well "indefinite" and hence not unique. So the antiderivative of a function is always expressed with a $+C$ (the constant of integration).
To make your reasoning precise we better get down to definite integrals. We then have $$\int_{1}^{x}t^{k}\,dt = \frac{x^{k + 1} - 1}{k + 1},k \neq -1,\,\int_{1}^{x} t^{-1}\,dt = \log x$$ Hence it is reasonable to expect that $$\lim_{k \to -1}\frac{x^{k + 1} - 1}{k + 1} = \log x$$ and yes this holds true when $x > 0$. Thus we may write $$\lim_{k \to -1}\int_{1}^{x}t^{k}\,dt = \int_{1}^{x}\left(\lim_{k \to -1}t^{k}\right)\,dt$$
Without using language of definite integrals you can argue in the following manner. As $k$ varies the function $x^{k}$ also varies (meaning that $x^{2}$ is a different function from $x^{3}$). Hence for our convenience we may choose the constant of integration involved to be different for each $k$. We thus choose $$\int x^{k}\,dx = \frac{x^{k + 1} - 1}{k + 1}$$ and then when $k \to -1$ we get $$\int \frac{dx}{x} = \log x$$ and the limit of $(x^{k + 1} - 1)/(k + 1)$ is $\log x$ when $k \to -1$.
