Find $f$ if $f'(x)=\dfrac{x^2-1}{x}$ knowing that $f(1) = \dfrac{1}{2}$ and $f(-1) = 0$ I am asked the following problem:

Find $f$ if $$f'(x)=\frac{x^2-1}{x}$$

I am not sure about my solution, which I will describe below:
My solution:
The first thing that I've done is separate the terms of $f'(x)$
\begin{align*}
f'(x)&=\frac{x^2-1}{x}\\
&=x-\frac{1}{x}\\
\therefore  \quad f(x)&=\frac{x^2}{2}-\ln|x|+c
\end{align*}
For ( x > 0 ):
\begin{align*}
f(x)&=\frac{x^2}{2}-\ln x+c\\
f(1)&=\frac{1^2}{2}-\ln 1+c=\frac{1}{2} \quad \Rightarrow \quad c=0\\
f(x)&=\frac{x^2}{2}-\ln |x|
\end{align*}
For ( x < 0 )
\begin{align*}
f(x)&=\frac{x^2}{2}-\ln (-x)+c\\
f(-1)&=\frac{(-1)^2}{2}-\ln [-(-1)]+c=0 \quad \Rightarrow \quad c=-\frac{1}{2}\\
f(x)&=\frac{x^2}{2}-\ln |x|-\frac{1}{2}
\end{align*}
Is my solution correct? Should I really find two different answers, one for $x > 0$ and another for $x < 0$?
Thank you.

Looks fine! However a little typo when calculate $f(-1)=0$ not $-1$
