Let $F$ a distribution function, $F(x)=a+b\tan^{-1}(\frac{x}{2})$ with $x\in\mathbb{R}$. Find $a,b$ and the density function. The distribution function of random variable $X$ is: 
$F(x)=a+b\tan^{-1}(\frac{x}{2})$ with $x\in\mathbb{R}$ Find $a,b$ and the density function.
My work:
Note $F$ is a continuous function in $\mathbb{R}$ then $X$ is a continuous random variable.
This implies, $F'(x)=f(x)$ where $f$ is density function.
Then, $f(x)=\frac{b}{2(1+(\frac{x}{2})^2)}=\frac{2b}{4+x^2}$ with $x\in \mathbb{R}$ is the density function
This implies $1=\int_{\infty}^{-\infty} \! f(x)  \,dx = btan^{-1}(\frac{x}{2})|^{\infty}_{-\infty}=b\pi\implies b=\frac{1}{\pi}$
Is good this result? Is good the reasoning?
I'm stuck with the value of $a$. Can someone help me?

Hint
Note that
$$
1=\lim_{x\to \infty}F(x)=a+b\frac{\pi}{2}
$$
and
$$
0=\lim_{x\to -\infty}F(x)=a-b\frac{\pi}{2}
$$
