Minus sign in logarithm of integral's solution I want to solve the following integral:
$
\int \frac{dp}{a(1-p)u-bp}
$
where $a$, $b$ and $u$ are some constants.

After integration I get:
$
p = -\frac{\log(-apu+au-bp)}{au+b} + C.
$

According to WolframAlpha it is equivalent to (citing precisely: "Which is equivalent for restricted $p$, $a$, $b$ and $u$ values to"):
$
p = -\frac{log(a(p-1)u+bp)}{au+b} + C,
$
or transforming it a bit, it is equivalent to:
$
p = -\frac{\log(apu-au+bp)}{au+b} + C.
$
So for some reasons a minus is "removed" from the logarithm. Could you please explain to me why?
Hint: compute $D(\log(-x))$ and compare with $D(\log(x))$.
\begin{align}
I &amp;= \int \frac{dp}{a(1-p)u - b p}  = \frac{1}{au} \, \int \frac{dp}{1 - \left( 1 + \frac{b}{au}\right) p } \\
&amp;= \frac{1}{au+b} \, \int \frac{dx}{1-x} \hspace{10mm} \text{where} \quad  x = \left(1+\frac{b}{au}\right) \, p \\
&amp;= - \frac{1}{au + b} \, \int \frac{dx}{x-1} \\
&amp;= - \frac{1}{au+b} \, \ln(x-1) \\
&amp;= - \frac{\ln(au\, (1-p) + bp) - \ln(au)}{au+b}.
\end{align}