Let $f(x)=\arctan x -\frac{ln|x|}{2}$. Then $f(x)$ is increasing in? Let $x<0$. $f'(x)=\frac{1}{1+x^2}+\frac{1}{2x}>0$
$$f'(x)=\frac{(1+x)^2}{2x(1+x^2)}$$ which is always negative. But answer is $x\in (-\infty,0)$.
 A: A:  The derivative of $\ln |x|=\frac 1 x$ for any $x\ne 0$
Proof:  for x<0,  $\ln |x|$=$\ln (-x)$  derivative by the chain rule is $\frac {du} {u}$, so $f'(x)=\frac {-1} {-x}=\frac 1 x$
So,  $f'(x)=\frac {1} {1+x^2} - \frac {1} {2x}$,  which when $x<0$, we have both terms are positive, so $f'(x)>0$ when $x<0$.  However, when $x>0$,  we need to consider which is bigger, $\frac {1} {1+x^2}$  or $\frac {1} {2x}$.  The larger fraction is going to be the one with the smaller denominator, so we can compare denominators to make our life easier.  consider $g(x)=1+x^2-2x=x^2-2x+1=(x-1)^2$.
This is always nonnegative, which means $1+x^2\ge 2x$. for $x\ge 0$], so $\frac {1} {1+x^2}\le \frac 1 {2x}$,  hence $f'(x)\le 0$ when $x>0$,  thus we only have increasing when $x\in (-\infty, 0)$
A: The derivative of $a(x)=|x|$ is 
$$
a'(x)=\frac{|x|}{x}
$$
(for $x\ne0$): indeed, $|x|/x=1$ for $x>0$ and $|x|/x=-1$ for $x<0$. So, by the chain rule, if $g(x)=\ln|x|$, we have
$$
g'(x)=\frac{1}{|x|}\frac{|x|}{x}=\frac{1}{x}
$$
This is a peculiarity of the logarithmic function; for $h(x)=\sin|x|$ we have
$$
h'(x)=\frac{|x|}{x}\cos|x|
$$
(for $x\ne0$).
So your function $f$ has, for $x\ne0$,
$$
f'(x)=\frac{1}{1+x^2}-\frac{1}{2x}=\frac{-1+2x-x^2}{2x(1+x^2)}
=-\frac{(x-1)^2}{2x(1+x^2)}
$$
which is clearly positive for $x<0$.
In the interval $(0,\infty)$ the derivative is negative, except at $1$, where $f'(1)=0$. Since the derivative vanishes only at an isolated point and is negative elsewhere, it is decreasing in the interval.
At $x=1$ there's a flex with horizontal tangent.

