Question

$\int \frac{1+{\tan }^{2}x}{1-{\tan }^{2}x}dx$ 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. $\frac{1}{2}\log \left(\frac{1+\tan x}{1-\tan x}\right)+c$

Answer 1

Answer: (D)

Let $I=\int \frac{1+{\tan }^{2}x}{1-{\tan }^{2}x}dx$
$=\int \frac{{\sec }^{2}x}{1-{\tan }^{2}x}dx$
Let $tandx=t$
$\Rightarrow$ $se{c}^{2}xdx=dt$
$\therefore$ $I=\int \frac{1}{1-t^2}dt$
$=\frac{1}{2}\int \left(\frac{1}{1-t}+\frac{1}{1+t}\right)dt$
$=\frac{1}{2}[-\log (1-t)+\log (1+t)]+c$
$=\frac{1}{2}\log \frac{1+t}{1-t}+c$
$=\frac{1}{2}\log \frac{1+\tan x}{1-\tan x}+c$