$\underset{8}{\overset{15}{\int }}\frac{dx}{(x+3)\sqrt{x+1}}$

$$\begin{gathered} {\int }_8^{15}\frac{\mathrm{dx}}{(\mathrm{x}+3)\sqrt{\mathrm{x}+1}} \\ ={\int }_3^4\frac{2\cancel{\mathrm{t} } \mathrm{dt}}{({\mathrm{t}}^2+2)\cancel{\mathrm{t} }} \\ ={\left[2 \frac{1}{\sqrt{2}} {\tan }^{-1}\left(\frac{\mathrm{t}}{\sqrt{2}}\right)\right]}_3^4 \\ =\sqrt{2}\left[{\tan }^{-1}\frac{4}{\sqrt{2}}-{\tan }^{-1}\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.\right] \\ =\sqrt{2} {\tan }^{-1}\frac{\left({\displaystyle \frac{4}{\sqrt{2}}}-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}{1+{\displaystyle \frac{12}{2}}} \\ =\sqrt{2} {\tan }^{-1}\left(\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}}{7}\right)=\sqrt{2} {\tan }^{-1} \frac{1}{7\sqrt{2}} \\ =\sqrt{2} {\cot }^{-1} 7\sqrt{2} \end{gathered}$$