This integral baffles me if anyone can help derive it, or demonstrate it, I truly appreciate it.
$$\int_0^ x\frac{\frac{t}{t^2+1}-\arctan(t)}{\arctan^2(t)+t^2} \,dt = \dfrac{\arctan\left(\dfrac{\arctan(x)}{x}\right)-\arctan \left(\dfrac{x}{\arctan(x)}\right)}{2} $$
A theorem? No, ERL.
Following achille hui's hint $$\frac{\frac{x}{1+x^2}-\arctan x}{(\arctan x)^2 + x^2} = \frac{x(\arctan x)' - \arctan x}{(\arctan x)^2 + x^2} = \frac{ x^2\left(\frac{\arctan x}{x}\right)'}{(\arctan x)^2 + x^2} = \frac{ \left(\frac{\arctan x}{x}\right)'}{1+\left(\frac{\arctan x}x\right)^2} $$ which makes $$I=\int\frac{\frac{x}{1+x^2}-\arctan x}{(\arctan x)^2 + x^2}\,dx=\int \frac{ \left(\frac{\arctan x}{x}\right)'}{1+\left(\frac{\arctan x}x\right)^2} \,dx$$ Now, the obvious change of variable $u=\frac{\arctan x}x$ leads to $$I=\int\frac{du}{1+u^2}=\arctan u=\arctan \left(\frac{\arctan x}{x}\right)$$ Then, provided $t\geq 0$,$$J=\int_0^t\frac{\frac{x}{1+x^2}-\arctan x}{(\arctan x)^2 + x^2}\,dx=\arctan \left(\frac{\arctan (t)}{t}\right)-\frac{\pi }{4}=\frac 12\left( 2\arctan \left(\frac{\arctan t}{t}\right)-\frac{\pi }{2}\right)$$ Now, use some classical trigonometric identities to get the rhs.