Why does this integral turn into -ln rather than +ln? \begin{align}
 &\int \frac{1}{100-y}\,\mathrm{d}y = \int x\,\mathrm{d}x\\
&\qquad\implies -\ln(100-y) = \frac{x^2}{2} + C_1.
\end{align}
Why does this integral turn into $-\ln$ rather than $+\ln$?
 A: "Why?" is kind of a nebulous question.  Generally speaking, the answer is "Because of the axioms of mathematics, and the resulting theorems."  However, that is clearly not what is being asked.  ;)  There are basically two approaches that can at least justify the sign.


*

*One attack is to implicitly differentiate each side of the second equation in order to obtain (via the chain rule)
$$
\frac{\mathrm{d}}{\mathrm{d}x} (-\ln(100-y)) = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{x^2}{2} + C_1 \right)
\implies -\frac{1}{100-y} \frac{\mathrm{d}}{\mathrm{d}x} (100-y) = \frac{1}{100-y}\frac{\mathrm{d}y}{\mathrm{d}x} = x,
$$
since $\frac{\mathrm{d}}{\mathrm{d}x} (100-y) = -1 \frac{\mathrm{d}}{\mathrm{d}x}$.  Using the standard jiggery-pokery of treating the differential like a variable and integrating, this becomes
$$
\frac{1}{100-y}\frac{\mathrm{d}y}{\mathrm{d}x} = x
\implies \frac{1}{100-y}\,\mathrm{d}y = x\,\mathrm{d}x
\implies \int\frac{1}{100-y}\,\mathrm{d}y = \int x\,\mathrm{d}x.
$$
Thus the result comes out of the chain rule.

*The other attack is to make a change of variables in the first equation.  Letting $u = 100-y$, we obtain $\mathrm{d}u = -\mathrm{d}y$.  Then
$$ \int \frac{1}{100-y}\,\mathrm{d}y
 = \int \frac{1}{u}\, -\mathrm{d}u
 = -\int \frac{1}{u}\, \mathrm{d}u
 = -\ln(u) + C
 = -\ln(100-y) + C.
$$
Note that I am playing a little fast and loose, and assuming that $100-y$ is positive over the domain of integration.  It might be better to write $\ln|u| = \ln|100-y|$, but this is a detail that can be resolved from context.  On the right, we obtain
$$ \int x\,\mathrm{d}x = \frac{x^2}{2} + C',$$
and so we conclude that
\begin{align}
\int \frac{1}{100-y}\,\mathrm{d}y = \int x\,\mathrm{d}x
&\implies \ln(100-y) + C = \frac{x^2}{2} + C' \\
&\implies \ln(100-y) = \frac{x^2}{2} + \underbrace{C+C'}_{=: C_1}. \end{align}
The two approaches are entirely equivalent, as the change of variables formula is only the chain rule applied to integrals via the Fundamental Theorem of Calculus.  However, the actual computations look a little different, thus it might be useful to see both.
I'll also note Cornman's "common rule" can be quickly shown to hold via a similar kind of change-of-variables argument.  The generalization that
$$ \int \frac{f'(x)}{f(x)} \,\mathrm{d}x = \ln|f(x)| + C $$
turns out to be quite useful (especially in complex analysis).  The adventurous soul might seek to work out the details of the rule.
