Solution:

Assume $I=\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$
Substituting $x+2=t \Rightarrow d x=d t$
On substituting the values of $x$ in given equation, we obtain
$\begin{array}{l}
\Rightarrow I=\int\left[2(t-2)^{2}+3\right] \sqrt{t} d t \\
\Rightarrow I=\int\left[2(t-2)^{2}+3\right] \sqrt{t} d t
\end{array}$
On expanding above equation using $(a-b)^{2}$ formula, we get
$\Rightarrow \mathrm{I}=\int\left[2 \mathrm{t}^{2}-8 \mathrm{t}+8+3\right] \sqrt{\mathrm{t}} \mathrm{dt}$
On simplification, we get
$\Rightarrow \mathrm{I}=\int\left[2 \mathrm{t}^{\frac{5}{2}}-8 \mathrm{t}^{\frac{3}{2}}+11 \mathrm{t}^{\frac{1}{2}}\right] \mathrm{dt}$
On integrating we obtain
$\begin{array}{l}
\Rightarrow I=\frac{4}{7} t^{\frac{7}{2}}-\frac{16}{5} t^{\frac{5}{2}}+\frac{22}{3} t^{\frac{3}{2}}+c \\
\Rightarrow I=\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c \\
\therefore \int\left(2 x^{2}+3\right) \sqrt{x+2} d x=\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c
\end{array}$