Answer
$F(x)=2x+3\cdot\tan^{-1}x+C$
Work Step by Step
$f(x)=\displaystyle \frac{2x^{2}+5}{x^{2}+1}$
$=\displaystyle \frac{2x^{2}+2+3}{x^{2}+1}$
$=\displaystyle \frac{2(x^{2}+1)}{x^{2}+1}+3\cdot\frac{1}{x^{2}+1}$
$=2+3\displaystyle \cdot\frac{1}{x^{2}+1}$
$=2x^{0}+3\displaystyle \cdot\frac{1}{x^{2}+1}$
From the Table of Antiderivatives,
$ f(x)=c\displaystyle \cdot x^{n}\Rightarrow F(x)=c\frac{x^{n+1}}{n+1},\quad$ and
$f(x)=\displaystyle \frac{1}{1+x^{2}} \Rightarrow F(x)=\tan^{-1}x$
So for $f(x)=2x^{0}+3\displaystyle \cdot\frac{1}{x^{2}+1}$,
$F(x)=2\displaystyle \cdot\frac{x^{0+1}}{0+1}+3\cdot\tan^{-1}x+C$
$F(x)=2x+3\cdot\tan^{-1}x+C$
