How to define the sign of a function $$y=\arctan\frac{x+1}{x-3} + \frac{x}{4}$$
I know that is necessary to put the function $>$ than $0$, but then?
$$\arctan\frac{x+1}{x-3} + \frac{x}{4}>0$$
It's a sum, so I can't set up a "false system" putting the two factors $>0$. In this case which is the rule to study the sign of this function?



Note: These are not homeworks.

Let $$f(x)=\arctan\frac{x+1}{x-3}+\frac{x}4\;.$$
You know that the algebraic sign of $\arctan u$ is the same as the sign of $u$, so 
$$
\arctan\frac{x+1}{x-3}\text{ is }\begin{cases}
\text{positive}&\text{when }x>3\\
\text{negative}&\text{when }-1<x<3\\
0&\text{when }x=-1\\
\text{positive}&\text{when }x<-1\;.
\end{cases}$$
Without further work we can be sure that $f(x)>0$ when $x>3$, and $f(x)<0$ when $-1\le x\le 0$; it’s only the intervals $(\leftarrow,-1)$ and $(0,3)$ that we need to investigate in detail.
Now
$$\begin{align*}
f\,'(x)&=\frac14-\frac4{(x-3)^2+(x+1)^2}\\
&=\frac{x^2-2x-3}{2\big((x-3)^2+(x+1)^2\big)}\\
&=\frac{(x-3)(x+1)}{2\big((x-3)^2+(x+1)^2\big)}
\end{align*}$$
is negative throughout the interval $0<x<3$, so $f$ is decreasing on that interval; since $f(0)<0$, it must be the case that $f(x)<0$ for all $x\in(0,3)$. Similarly, $f\,'(x)>0$ when $x<-1$, so $f$ is increasing on the interval $(\leftarrow,-1)$, and $f(-1)=0$, so $f(x)<0$ for all $x<-1$. To summarize,
$$f(x)\text{ is }\begin{cases}
\text{positive}&\text{when }x>3\\
\text{negative}&\text{when }x<3\;.
\end{cases}$$
I don’t at the moment see any simple way to avoid using some calculus here. You can rewrite
$$\frac{x+1}{x-3}=1+\frac4{x-3}$$
to see easily that $\frac{x+1}{x-3}<1$ when $x<3$, so $0<\arctan\frac{x+1}{x-3}<\frac{\pi}4$, and therefore $f(x)<0$ when $x\le-\pi$, but that still leaves the intervals $(-\pi,-1)$ and $(0,3)$ to be dealt with.
