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) \[\frac{1}{2}\log \left( \frac{1+\tan x}{1-\tan x} \right)+c\]
Correct Answer: D
Solution :
\[I=\int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}dx}\]\[=\int{\frac{{{\sec }^{2}}x}{1-{{\tan }^{2}}x}dx}\] Put \[\tan x=t\] Þ \[{{\sec }^{2}}x.\,dx=dt\] Þ \[I=\int{\frac{dt}{1-{{t}^{2}}}}\] \[=\frac{1}{2\times 1}\log \left[ \frac{1+t}{1-t} \right]+c\]\[=\frac{1}{2}\log \left| \frac{1+\tan x}{1-\tan x} \right|+c\].
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