A) \[\tan (\log x)+c\]
B) \[\log (\sec x)+c\]
C) \[\log (\tan x)+c\]
D) \[\sec (\log x)\ .\ \tan (\log x)+c\]
Correct Answer: A
Solution :
Put \[t=\log x\Rightarrow x\,dt=dx,\] then \[\int_{{}}^{{}}{\frac{1}{x}{{\sec }^{2}}(\log x)\,dx}=\int_{{}}^{{}}{{{\sec }^{2}}t\,dt=\tan t+c}=\tan (\log x)+c\].You need to login to perform this action.
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