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