Answer
$F\left( x \right) = 3\left( {{{\tan }^{ - 1}}x} \right) + 2x + C$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{2{x^2} + 5}}{{{x^2} + 1}} \cr
& {\text{Using the long division we obtain }}\frac{{2{x^2} + 5}}{{{x^2} + 1}} = \frac{3}{{{x^2} + 1}} + 2,{\text{ so}} \cr
& f\left( x \right) = \frac{3}{{{x^2} + 1}} + 2 \cr
& f\left( x \right) = 3\left( {\frac{1}{{{x^2} + 1}}} \right) + 2 \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}\frac{1}{{1 + {x^2}}} \to {\text{Particular antiderivative: ta}}{{\text{n}}^{ - 1}}x \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& f\left( x \right) = 3\left( {\frac{1}{{{x^2} + 1}}} \right) + 2 \cr
& F\left( x \right) = 3\left( {{{\tan }^{ - 1}}x} \right) + 2x + C \cr} $$