$\text{ Integral of }\frac{{\sec }^{2}xdx}{3{\tan }^{2}x+5{\sec }^{2}x+1}dx\text{ is equal to }$

$\int \frac{{\sec }^{2}xdx}{3{\tan }^{2}x+5{\sec }^{2}x+1}=\int \frac{{\sec }^{2}xdx}{3{\tan }^{2}x+5\left(1+{\tan }^{2}x\right)+1}$

$\tan x=r$

$\begin{array}{r}{\sec }^{2}xdx=dr\\ =\int \frac{dr}{3r^2+5r^2+6}\end{array}$

$\begin{array}{l}=\int \frac{dr}{8r^2+6}\\ =\frac{1}{8}\int \frac{dr}{r^2+\frac{6}{8}}\\ =\frac{1}{8}\int \frac{dr}{r^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\end{array}$

$\text{ Use formula }I=\int \frac{1}{r^2+a^2}dr$

$\begin{array}{l}=\frac{1}{a}{\tan }^{-1}\left(\frac{r}{a}\right)+k\\ =\frac{1}{8}\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{r}{\sqrt{3}}2\right)+k\\ =\frac{1}{4\sqrt{3}}{\tan }^{-1}\left(\frac{2r}{\sqrt{3}}\right)+k\end{array}$