Integral of arctan form I am trying to solve this integral without techniques ($u$-sub/parts), just simplifying and inspection:
$$
\int \dfrac{1}{4+x^{2}}dx
$$
I notice an $\arctan$ form, but that $4$ in the denominator is confusing me, if there was a multiplying factor in front of the $x$ I would know what to do, but stumped on this case.

One can take a different route with the following. Let $x = 2 t$ to obtain
\begin{align}
I &= \int \frac{dx}{4 + x^2} \\
&= \frac{1}{2} \, \int\frac{dt}{1 + t^2} = \frac{1}{4} \, \int\left(\frac{1}{1 + i t} + \frac{1}{1 - i t} \right) \, dt \\
&= \frac{1}{4} \, \left[ \frac{1}{i} \, \ln(1 + i t) - \frac{1}{i} \, \ln(1-i t)  \right] + c_{1} \\
& = \frac{1}{4i} \, \ln\left(\frac{1+i t}{1- i t}\right) + c_{1}.
\end{align}
Now using 
$$\tan^{-1}(y) = \frac{1}{2i} \, \ln\left(\frac{1+ i y}{1 - i y}\right)$$
then the integral becomes, after back substitution,
$$\int \frac{dx}{4 + x^2} = \frac{1}{2} \, \tan^{-1}\left(\frac{x}{2}\right) + c_{1} = \frac{1}{4i} \, \ln\left(\frac{2 +i x}{2 - i x}\right) + c_{0}.$$
