Integral Calculus Question 168
Question: $ \int_{{}}^{{}}{\frac{1+\tan x}{x+\log \sec x}\ dx=} $
[AI CBSE 1986]
Options:
A) $ \log (x+\log \sec x)+c $
B) $ -\log (x+\log \sec x)+c $
C) $ \log (x-\log \sec x)+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ t=x+\log \sec x\Rightarrow dt=(1+\tan x),dx, $ then $ \int_{{}}^{{}}{\frac{1+\tan x}{x+\log \sec x},dx=\int_{{}}^{{}}{\frac{1}{t},dt=\log t+c}} $ $ =\log (x+\log \sec x)+c. $
BETA
