Ta có:
$\begin{array}{l}
\int\limits_0^3 {\frac{{{x^2} + 1}}{{\sqrt {x + 1} }}dx} = \int\limits_0^3 {\frac{{\left(
{{x^2} - 1} \right) + 2}}{{\sqrt {x + 1} }}dx = \int\limits_0^3 {\left[ {(x - 1)\sqrt {x + 1} +
\frac{2}{{\sqrt {x + 1} }}} \right]dx} } \\
= \int\limits_0^3 {\left[ {\left( {(x + 1) - 2} \right)\sqrt {x + 1} } \right]dx} + 2\int\limits_0^3
{\frac{1}{{\sqrt {x + 1} }}dx} \end{array}$
$= \frac{2}{5}\sqrt{(x+1)^5}|^3_0-\frac{4}{3}\sqrt{(x+1)^3}|^3_0+4\sqrt{x+1}|^3_0$
$\begin{array}{l}=\frac{2}{5}
(32-1)-\frac{4}{3}(8-1)+4=\frac{62}{5}-\frac{28}{3}+4=\frac{{106}}{{15}}
\end{array}$