$X$ and $Y$ are independent rv having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$ determine pdf of $Z=\frac{X+Y}{3}$ 
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.

So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$. 
We shall find the CDF of $Z$ and then differentiate it.
$F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \  dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.
How do I do this? Or otherwise is there different more elegant way to get the required pdf?

If you are comfortable with performing a slightly hefty partial fraction decomposition while evaluating an integral, you could do the following standard procedure:
Denote the joint density of $(X,Y)$ as $f_{X,Y}(x,y)$ where the support of the distribution is $x,y\in\mathbb{R}$
Transform $(X,Y)\to (U,V)$ such that $U=\frac{X+Y}{3}, V=Y$
$\Rightarrow x=3u-v,\quad y=v$ where $u,v\in\mathbb R$
The Jacobian of the transformation is 
$\det\left(J\left(\dfrac{x,y}{u,v}\right)\right)=\det\begin{pmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\\\end{pmatrix}=\det\begin{pmatrix}3&-1\\ 0& 1\end{pmatrix}=3$
The joint density of $(U,V)$ is then given by
$f_{U,V}(u,v)=f_{X,Y}(3u-v,v)|\det(J)|=\dfrac{3}{\pi^2(1+(3u-v)^2)(1+v^2)},\qquad u,v\in\mathbb{R}$
So the marginal density of $U$ is 
$f_U(u)=\displaystyle\int_\mathbb Rf_{U,V}(u,v)\,\mathrm{d}v$
$\qquad\quad=\dfrac{3}{\pi^2}\displaystyle\int_\mathbb R\frac{\mathrm{d}v}{(1+(3u-v)^2)(1+v^2)}$
$\qquad\quad=\dfrac{3}{\pi^2}\cdot\dfrac{2\pi}{9u^2+4}=\dfrac{6}{\pi(9u^2+4)}\quad,u\in\mathbb{R}$
You can refer to this post for evaluating the above integral.
