Let $I=\int\frac{5x+3}{\sqrt{x^2+4x+10}}dx$ Put $5x+3=A\left(\frac{d}{dx}\left(x^2+4x+10\right)\right)+B$ $\Rightarrow 5x+3=A(2x+4)+B$ ... (i) Comparing the coefficients of x in (i), we get $5=2A\Rightarrow A=\frac{5}{2}$ Comparing the constant terms in (i), we get $3=4A+B\Rightarrow B=-7$ $I=\int\frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}}dx$$I=\frac{5}{2}I_1-7I_2\text{ (say)}$ ... (ii)Now, $I_1=\int\frac{2x+4}{\sqrt{x^2+4x+10}}dx$ Put $x^2+4x+10=t$ $\Rightarrow (2x+4)dx=dt$ $I_1=\int\frac{dt}{\sqrt{t}}=2\sqrt{t}$ $=2\sqrt{x^2+4x+10}+C_1$ ... (iii) and $I_2=\int\frac{dx}{\sqrt{x^2+4x+10}}$ $=\frac{dx}{\sqrt{(x+2)^2+(\sqrt{6})^2}}$ $=\log\left|x+2+\sqrt{(x+2)^2+(\sqrt{6})^2}\right|$$=\log\left|x+2+\sqrt{x^2+4x+10}\right|+C_2$... (iv)Hence from (ii), (iii) and (iv), we get