When should the antiderivative of a rational function be defined as a piecewise function? I'm doing basic problems on antiderivatives and there seems to be an inconsistency in my book.
The instructions for these problems are:

Find the most general antiderivative of the function.

Number 11 is:

11. $f(x) = \dfrac{10}{x^9}$

So, I naturally wrote down $F(x) = -\dfrac{5}{4x^8} + C$.
The solution manual says this is wrong.  The function has domain $(-\infty, 0) \cup (0, \infty)$, so $F(x) = \begin{cases} -\dfrac{5}{4x^8}+C_1 & \text{if } x < 0 \\
-\dfrac{5}{4x^8}+C_2 & \text{if } x > 0 \end{cases}$
Okay, I thought.  That makes sense.  
I then come to number 13, which is:

13. $f(x) = \dfrac{u^4 + 3\sqrt{u}}{u^2}$.

I figured the domain is $(-\infty, 0) \cup (0, \infty)$, so I wrote down $F(x) = \begin{cases} \frac{1}{3}u^3-6u^{-1/2}+C_1 & \text{if } u > 0 \\
\frac{1}{3}u^3-6u^{-1/2}+C_2 & \text{if } u < 0 \end{cases}$.
Well, the solution book doesn't mention the domain and just says $F(x) = \frac{1}{3}u^3-6u^{-1/2}+C$.
I later realized that the $\sqrt{u}$ in the numerator and the $u^2$ in the denominator must limit the domain to the positive numbers, so the antiderivative doesn't need to be defined for anything but positive numbers.  So, okay, I think I get it.
Next number 19 is:

19. $f(x) = \dfrac{x^5-x^3+2x}{x^4} = x - \dfrac{1}{x} + \dfrac{2}{x^3}$

So, again, since the domain appears to be $(-\infty, 0) \cup (0, \infty)$, I wrote down $F(x) = \begin{cases} \frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C_1 & \text{if } x > 0 \\
\frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C_2 & \text{if } x < 0 \end{cases}$.
But the book again doesn't mention the domain and just says that $F(x) = \frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C$.
I am confused.  This doesn't seem consistent, especially between #11 and #19.  Why is my answer not right for #19?

The person who wrote out the solution may have thought that putting absolute value signs around $x$ takes care of the problem. It doesn't. Congratulations on reading with care.
The answer for Question $19$ ought to be something like
$$F(x) = \begin{cases} \frac{1}{2}x^2 - \ln x - \dfrac{1}{x^2} + C_1 & \text{if } x > 0 \\
\frac{1}{2}x^2 - \ln(-x) - \dfrac{1}{x^2} + C_2 & \text{if } x < 0 \end{cases}.$$
