Let $\mathrm{f}\left(\mathrm{x}\right)=\int \frac{2\mathrm{x}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+3)}\mathrm{dx}$ . If $\mathrm{f}\left(3\right)=\frac{1}{2}({\log }_{\mathrm{e}}5-{\log }_{\mathrm{e}}6)$ , then  f(4) is equal to 

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\int \frac{2\mathrm{x}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+3)}\mathrm{dx}\\ \mathrm{Let}\mathrm{\hspace{0.17em}}{\mathrm{x}}^2=\mathrm{t}\\ 2\mathrm{xdx}=\mathrm{dt}\\ \int \frac{\mathrm{dt}}{(\mathrm{t}+1)(\mathrm{t}+3)}\\ =\frac{1}{2}\int \frac{(\mathrm{t}+3)-(\mathrm{t}+1)}{(\mathrm{t}+1)(\mathrm{t}+3)}\mathrm{dt}\\ =\frac{1}{2}[\ln |\mathrm{t}+1|-\ln |\mathrm{t}+3|]+\frac{\mathrm{C}}{2}\\ =\frac{1}{2}[\ln |{\mathrm{x}}^2+1|-\ln |{\mathrm{x}}^2+3|]+\frac{\mathrm{C}}{2}\\ \therefore \mathrm{f}\left(3\right)=\frac{1}{2}[\ln 5-\ln 6]\\ \therefore \frac{1}{2}[\ln 5-\ln 6]=\frac{1}{2}[\ln 10-\ln 12]+\frac{\mathrm{C}}{2}\\ \Rightarrow \mathrm{C}=0\\ \therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}[\ln |{\mathrm{x}}^2+1|-\ln |{\mathrm{x}}^2+3|]\\ \mathrm{f}\left(4\right)=\frac{1}{2}[\ln 17-\ln 19]\end{array}$