How to find the slant asymptote for $y=\frac{x^2\arctan\left(x\right)}{3x+3}$ I got stuck at the end and wonder if you can give some guide on how to proceed to find the slant asymptote?
$$y=\frac{x^2\arctan\left(x\right)}{3x+3}$$

$$y= kx+m\\
\frac{f(x)}{x} \rightarrow k\\
f(x) - kx \rightarrow m $$ 

Solution:
$$\lim _{x\to \infty }\left(\frac{\frac{x^2\arctan\left(x\right)}{3x+3}}{x}\:\right)=\lim _{x\to \infty }\left(\frac{x^2\arctan\left(x\right)}{x(3x+3)}\:\right)=\lim _{x\to \infty }\left(\frac{x^2\arctan\left(x\right)}{(3x^{2\:\:}+3x)}\:\right)\\
=\lim _{x\to \infty }\left(\frac{x^2}{x^2} \cdot\frac{\arctan\left(x\right)}{(3+\frac{3}{x})}\:\right)=\frac{\pi}{6}\Longrightarrow k=\frac{\pi}{6}\\
f(x)-kx \implies \frac{x^2\arctan\left(x\right)}{3x+3} - \frac{\pi}{6} \cdot {x}\\
\frac{6\cdot x^2\arctan -3\pi{x^2} - 3\pi {x}}{18(x+1)}$$
Searching for $m$.

hint
Put $y=\frac 1x$ and use the fact that
for $y>0$, we have
$$\arctan(\frac 1y)=\frac {\pi}{2}-\arctan(y)$$
use L'Hopital rule to compute the limit when $y\to 0^+$.
