Integrate the function: $\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$

Let $5 x+3=A \frac{d}{d x}(x^{2}+4 x+10)+B$
$\Rightarrow 5 x+3=A(2 x+4)+B$
Equating the coefficients of $x$ and the constant term,we obtain:
$2 A=5 \Rightarrow A=\frac{5}{2}$
$4 A+B=3 \Rightarrow B=3-4(\frac{5}{2})=3-10=-7$
$\therefore 5 x+3=\frac{5}{2}(2 x+4)-7$
$\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{(x+2)^{2}+6}} d x$
For the first integral,let $t=x^{2}+4 x+10$,then $dt=(2 x+4)dx$. Thus,$\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^{2}+4 x+10}$.
For the second integral,using $\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=\log |x+\sqrt{x^{2}+a^{2}}|$,we get $\log |(x+2)+\sqrt{(x+2)^{2}+6}| = \log |(x+2)+\sqrt{x^{2}+4 x+10}|$.
Combining these,the result is $5\sqrt{x^{2}+4 x+10}-7 \log |(x+2)+\sqrt{x^{2}+4 x+10}|+C$.