Find : $\displaystyle \int \frac{5 \mathrm{x}}{(\mathrm{x}+1)\left(\mathrm{x}^{2}+9\right)} \mathrm{d} \mathrm{x}$.

\[\begin{aligned} & \frac{5 \mathrm{x}}{(\mathrm{x}+1)\left(\mathrm{x}^{2}+9\right)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+9} \\ & \Rightarrow \mathrm{~A}=-\frac{1}{2}, \mathrm{~B}=\frac{1}{2}, \mathrm{C}=\frac{9}{2} \end{aligned} \] Given integral \[\begin{aligned} & =-\frac{1}{2} \int \frac{1}{\mathrm{x}+1} \mathrm{d} \mathrm{x}+\frac{1}{2} \int \frac{\mathrm{x}+9}{\mathrm{x}^{2}+9} \mathrm{d} \mathrm{x} \\ & =-\frac{1}{2} \int \frac{1}{\mathrm{x}+1} \mathrm{d} \mathrm{x}+\frac{1}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^{2}+9} \mathrm{d} \mathrm{x}+\frac{1}{4} \int \frac{18}{\mathrm{x}^{2}+9} \mathrm{d} \mathrm{x} \\ & =-\frac{1}{2} \log |\mathrm{x}+1|+\frac{1}{4} \log \left(\mathrm{x}^{2}+9\right)+\frac{3}{2} \tan ^{-1} \frac{\mathrm{x}}{3}+\mathrm{C} \end{aligned} \]