MCQ
$\int_{}^{} {\frac{1}{x}{{\sec }^2}(\log x)dx = } $
  • $\tan (\log x) + c$
  • B
    $\log (\sec x) + c$
  • C
    $\log (\tan x) + c$
  • D
    $\sec (\log x)\;.\;\tan (\log x) + c$

Answer

Correct option: A.
$\tan (\log x) + c$
a
(a) Put $t = \log x \Rightarrow x\,dt = dx,$ then
$\int_{}^{} {\frac{1}{x}{{\sec }^2}(\log x)\,dx} = \int_{}^{} {{{\sec }^2}t\,dt = \tan t + c} = \tan (\log x) + c$.

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