$\textsf{evaluate}$

\begin{align}

\displaystyle

{I}&={\int{\frac{x+2}{x^2+1}dx}}\\

&=\int{\frac{x}{x^{2}{+1}}dx{\ +\ 2}\int{\frac{1}{x^{2}{+1}}}}{\ }dx\\

u&=x^{2}+1 \therefore \frac{1}{2x}du=dx\\

x&=\sqrt{u-1}\\

\end{align}

...?

$\textit{calculator answer.?}$

$\dfrac{\ln\left(x^2+1\right)}{2}

+2\arctan\left(x\right)+C$