Why is $\int \frac 1u du = \ln u$? Why, in $u$-substitution if $u$ appears to the $-1$ power it becomes equivalent to the $\ln u$? 
I don't have a specific problem where that is the case, but I do recall this being a rule of thumb. Can someone explain this to me?
 A: The true answer comes from real analysis as Alex explained above. But I suspect you want a more intuitive answer. If $\int \frac{1}{u}\,du = \ln u$, this is basically the same as saying that $\frac{d}{du}\ln u = \frac{1}{u}$.
Recall that $\ln u$ is the inverse function of $e^u$, and you might have come across this theorem: given sufficient differentiability conditions, $\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$. If $f(u) = e^u$, then $f^{-1}(u) = \ln u$ and $\frac{d}{du}e^u = e^u$, so $\frac{d}{du}f^{-1}(u) = \frac{1}{f'(f^{-1}(u))} = \frac{1}{e^{\ln u}} = \frac{1}{u}$.
This isn't a good proof by mathematical standards, but perhaps it shows you where this is coming from.
A: \begin{align}
\int_1^xu^{-1}\operatorname d\!u&=\lim_{\epsilon\to0}\int_1^xu^{\epsilon-1}\operatorname d\!u\\
&=\lim_{\epsilon\to0}\left(\frac{x^\epsilon}\epsilon-\frac{1^{\epsilon}}\epsilon\right)\\
&=\lim_{\epsilon\to0}\frac{x^\epsilon-1}\epsilon\\
&(\rm{L'H\hat opital\ with\ respect\ to\ }\epsilon)\\
&=\lim_{\epsilon\to0}\frac{(\ln x)x^\epsilon-0}1\\
&=\ln x
\end{align}
A: The reason the $\int_1^x \frac 1u du= \ln x$ is because that's simply what it's defined to be. Then the fundamental theorem of calculus tells us that the antiderivative must be the integrand and so $\int \frac 1xdx = \ln x +C$. The natural logarithm was originally defined as this integral and all of its properties fall directly from this and the fundamental theorem of calculus, for example it should be extremely clear now why $\ln 0$ is undefined! If you start learning more about real analysis you'll learn all about it!  
