Integral Calculus Question 475
Question: $ \int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x},dx} $ equals to
[RPET 2001]
Options:
A) $ \log ( \frac{1-\tan x}{1+\tan x} )+c $
B) $ \log ( \frac{1+\tan x}{1-\tan x} )+c $
C) $ \frac{1}{2}\log ( \frac{1-\tan x}{1+\tan x} )+c $
D) $ \frac{1}{2}\log ( \frac{1+\tan x}{1-\tan x} )+c $
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Answer:
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 [ \frac{1+t}{1-t} ]+c $ $ =\frac{1}{2}\log | \frac{1+\tan x}{1-\tan x} |+c $ .
BETA
