\[\int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}\,dx}\] equals to [RPET 2001]

\[\log \left( \frac{1-\tan x}{1+\tan x} \right)+c\]

\[\log \left( \frac{1+\tan x}{1-\tan x} \right)+c\]

\[\frac{1}{2}\log \left( \frac{1-\tan x}{1+\tan x} \right)+c\]

\[\frac{1}{2}\log \left( \frac{1+\tan x}{1-\tan x} \right)+c\]

\[\int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}\,dx}\] equals to

1.	\[\int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}\,dx}\] equals to [RPET 2001]
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» E.

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\[\int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}\,dx}\] equals to [RPET 2001]

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