<s>[INST] Evaluate $\int_{1}^{4} \int_{-1}^{2z} \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy \; dx\; dz$ $\int_{1}^{4} \int_{-1}^{2z} \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy \; dx\; dz$
Consider $ \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy =  $ $\int_{0}^{\sqrt{3}x} \frac{x}{x^2 +y^2} - \frac{y}{x^2+y^2} \;dy$
$\int_{0}^{\sqrt{3}x} \frac{x}{x^2 +y^2} \; dy  = \arctan(\sqrt{3}) = \pi/3$
$x^2 + y^2 = t \Rightarrow y \; dy = dt/2$
$ \int_{0}^{\sqrt{3}x} \frac{y}{x^2 +y^2} \;dy = \int_{x^2}^{3+x^2} \frac{dt}{t} = \frac{\ln(3+x^2)}{2} - \ln(x)$
$ \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy  = \pi/3 + \ln(x) - \frac{\ln(3+x^2)}{2}$ 
Now it will be very lengthy and  difficult to integrate this with respect to x and z.
 So I am stuck here. Is there any easy method to solve such questions ?
 [/INST] $$\int\frac{x-y}{x^2+y^2}\,dy=\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\log(x^2+y^2).$$
Therefore, your integral becomes
$$\begin{align}&\int_1^4\int_{-1}^{2z}\left(\arctan(\sqrt{3})-\frac{1}{2}\log(4x^2)-\arctan0+\frac{1}{2}\log(x^2)\right)\,dx\,dz\\ 
=&\int_1^4\int_{-1}^{2z}\left(\arctan(\sqrt{3})-\frac{1}{2}\log(4)-\frac{1}{2}\log(x^2)-\arctan0+\frac{1}{2}\log(x^2)\right)\,dx\,dz\\
=&\int_1^4\int_{-1}^{2z}\left(\frac{\pi}{3}-\log(2)\right)\,dx\,dz\\
=&\left(\frac{\pi}{3}-\log(2)\right)\int_1^4(2z+1)dz\\
=&18\left(\frac{\pi}{3}-\log(2)\right)\\
=&6\pi-18\log 2
\end{align}$$
Your mistake was in calculating 
$\int\frac{y}{x^2+y^2}\,dy.$ Note that
$$\int\frac{y}{x^2+y^2}\,dy=\frac{1}{2}\log(x^2+y^2).$$
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