Integral Calculus Question 477
Question: $ \int_{{}}^{{}}{\frac{1}{x}{{\sec }^{2}}(\log x)dx=} $
Options:
A) $ \tan (\log x)+c $
B) $ \log (\sec x)+c $
C) $ \log (\tan x)+c $
D) $ \sec (\log x)\ .\ \tan (\log x)+c $
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Answer:
Correct Answer: A
Solution:
Put $ t=\log x\Rightarrow x,dt=dx, $ then $ \int_{{}}^{{}}{\frac{1}{x}{{\sec }^{2}}(\log x),dx}=\int_{{}}^{{}}{{{\sec }^{2}}t,dt=\tan t+c}=\tan (\log x)+c $ .
BETA
