Q. $\int_{}^{}{\frac{1 + \tan^{2}x}{1 - \tan^{2}x}\ dx}$ is equal to
A) $\log{\left( \frac{1 - \tan x}{1 + \tan x} \right) + c}$
B) $\log{\left( \frac{1 + \tan x}{1 - \tan x} \right) + c}$
C) $\frac{1}{2}\log{\left( \frac{1 - \tan x}{1 + \tan x} \right) + c}$
D) $\frac{1}{2}\log{\left( \frac{1 + \tan x}{1 - \tan x} \right) + c}$
Explanation
(d)
Let $I = \int_{}^{}{\frac{1 + \tan^{2}x}{1 - \tan^{2}x}\ dx} = \int_{}^{}{\frac{\sec^{2}x}{1 - \tan^{2}x}\ dx}$
Put $\tan x = t\ \ \ \Rightarrow \sec^{2}xdx = dt$
$\therefore I = \int_{}^{}\frac{dt}{1 - t^{2}} = \frac{1}{2 \times 1}\log\left( \frac{1 + t}{1 - t} \right) + c$
$= \frac{1}{2}\log{\left( \frac{1 + \tan x}{1 - \tan x} \right) + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.