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?