$\mathrm{I}=\int \frac{\frac{5}{2}(2 x+4)+(3-10)}{\sqrt{x^2+4 x+10}} d x$
$=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^2+4 x+10}} d x-7 \int \frac{d x}{\sqrt{x^2+4 x+10}}$
$=\frac{5}{2} \mathrm{I}_1-7 \mathrm{I}_2+\mathrm{C}$ (say)
Put $x^2+4 x+10=\mathrm{t}, \Rightarrow(2 x+4) \mathrm{d} x=\mathrm{dt}$
$\therefore \mathrm{I}_1=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{x^2+4 x+10}$
$\mathrm{I}_2=\int \frac{d x}{\sqrt{(x+2)^2+(\sqrt{6})^2}}$
$=\log \left|x+2+\sqrt{x^2+4 x+10}\right|$
$\mathrm{I}=5 \sqrt{x^2+4 x+10}-7 \log \mid x+2+\sqrt{x^2+4 x+10 \mid}+\mathrm{C}$