<s>[INST] When is y increasing? Arctan is the inverse of tan.
I tried to normally differentiate and solve for y' but the LCM is complicated.
Is there any easy substitution simplifies the problem?
$$y=(\frac{1}{3})\textrm{log}(\frac{x+1}{\sqrt{x^2-x+1}}) + \frac{1}{\sqrt{3}}\textrm{arctan}(\frac{2x+1}{\sqrt3})$$
 [/INST] $$y=\frac{1}{3}\ln{\frac{x+1}{\sqrt{x^2-x+1}}}+\frac{1}{\sqrt{3}}\tan^{-1}{\frac{2x+1}{\sqrt{3}}}$$
Before differentiating, we will manipulate the function to make it easier to differentiate.
$$y=\frac{\ln(x+1)}{3}-\frac{\ln(x^2-x+1)}{6}+\frac{1}{a}\tan^{-1}\frac{u}{a}$$where $a=\sqrt3$
$$\frac{\delta y}{\delta x}=\frac{1}{3(x+1)}-\frac{2x-1}{6(x^2-x+1)}+\frac{\delta u}{a^2+u^2}$$
The derivative of the inverse tangent part of the function follows by reversing the standard integration tables.
$$\frac{\delta y}{\delta x}=\frac{2x^2-2x+2-2x^2-2x+x+1}{6(1+x^3)}+\frac{2}{4x^2+4x+1+3}=\frac{-3x+3}{6(1+x^3)}+\frac{1}{2(x^2+x+1)}=\frac{1-x}{2(1+x^3)}+\frac{1}{2(x^2+x+1)}=\frac{1-x^3+1+x^3}{2(1+x^3)(1+x+x^2)}$$
$$\frac{\delta y}{\delta x}=\frac{1}{(1+x)(1-x+x^2)(1+x+x^2)}$$
The only term which has a change in sign is $(1+x)$, the other two quadratic factors are positive for all real $x$.
Hence, your function has a positive gradient for all $x>-1$ and therefore, it is strictly increasing in its domain for all $x>-1$. We do not take the equality, since that is a singularity of the function.
</s>