Question

$\int \frac{{\sec }^2\mathrm{x}}{1-{\tan }^{2}x}dx$

Answer 1

$$\begin{gathered} \int \frac{{\sec }^{2}xdx}{1-{\tan }^{2}x} \\ \mathrm{let}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \mathrm{Now},\int \frac{{\sec }^{2}xdx}{1-{\tan }^{2}x} \\ =\int \frac{dt}{1-t^2} \\ =\frac{1}{2}\log \left|\frac{1+t}{1-t}\right|+C \\ =\frac{1}{2}\log \left|\frac{1+\tan x}{1-\tan x}\right|+C \end{gathered}$$