Integral Calculus Question 257
Question: $ \int_{{}}^{{}}{\frac{1}{x{{\cos }^{2}}(1+\log x)}\ dx=} $
Options:
A) $ \tan ,(1+\log x)+c $
B) $ \cot ,(1+\log x)+c $
C) $ -\tan ,(1+\log x)+c $
D) $ -\cot (,1+\log x)+c $
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ 1+\log x=t\Rightarrow \frac{1}{x},dx=dt, $ then $ \int_{{}}^{{}}{\frac{1}{x{{\cos }^{2}}(1+\log x)},dx}=\int_{{}}^{{}}{\frac{dt}{{{\cos }^{2}}t}=\int_{{}}^{{}}{{{\sec }^{2}}t,dt}} $ $ =\tan t+c=\tan (1+\log x)+c. $
BETA
