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 $ tan\text{ }dx=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 $