A question about the absolute value in integrals I do really understand why we put the absolute value when integrating functions leading to $\log$ function for example:
$$ \int{\dfrac{\mathrm dx}x}=\log\lvert x\rvert + C$$
, it is very common in books to see them ignoring the absolute value (this is possible of course under some assumptions), but sometimes I do not see any assumptions. for example the following integral :
 $$\int \dfrac{du}{1-u}= \int dt$$
the expected result is: $$\log\lvert 1-u\rvert = -t + C $$
but Maple gives the result without the absolute value (I did not used the assume function)
EDIT 1
I want really to know in general case if the absolute value is required or it does not matter even if there are no assumptions
I am so confused, any ideas please?
 A: In calculus courses one deals with the function $\ln\colon(0,+\infty)\to\mathbb R$. For $x>0$ it holds that
$$
\frac{d}{dx}\ln x=\frac{1}{x}.
$$
Note that the function $x\mapsto 1/x$ is defined for $x\in\mathbb R$ with $x\neq 0$. Integrating, one has
$$
\int \frac{1}{x}\,dx=
\begin{cases}
\ln x+C_1 & x>0\\
\ln (-x)+C_2 & x<0.
\end{cases}
\tag{*}
$$
This is sometimes written
$$
\int\frac{1}{x}\,dx=\ln |x|+C
$$
but in applications one consider either $x>0$ or $x<0$.
In complex analysis one also deals with a logarithm, taking complex numbers as arguments (I will not dig into the subject of branches here). One common definition is
$$
\log z=\ln |z|+i\arg z.
$$
Here $\ln$ is the logarithm defined in calculus. $\arg$ is the argument function. As it happens, also
$$
\frac{d}{dz}\log z=\frac{1}{z}.
$$
Here $\frac{d}{dz}$ is the complex derivative. Thus, in the complex setting
$$
\int\frac{1}{z}\,dz=\log z+C,
$$
i.e. the logarithm is still an antiderivative of $1/z$. Here, again, one should be more careful with domains, but since I understand it as OP is not into complex analysis yet (correct me if wrong), I do not dwell about that.
Finally, maple has no reason to expect the $x$ in $1/x$ to be real, so when you hit
 int(1/x,x);

maple makes no assumption on $x$, but return

$\log x$.

Note that, if $x$ happens to be real and positive, and one assumes $\arg x=0$ (which is natural), then
$$
\log x=\ln|x|+i\arg x=\ln x
$$
so the different logarithms really coincide.
What should you do?
I suggest that you, every time you must integrate something that will return a logarithm, pause and think if you know anything about the variable. If you know that $x$ is positive, then $\int 1/x\,dx=\ln x+C$. If you know that $x$ is negative, then $\int 1/x\,dx=\ln(-x)+C=\ln|x|+C$ and if you do not know, then use $(*)$ above.
