MCQ
$\int_{}^{} {\frac{{1 + \tan x}}{{x + \log \sec x}}\;dx = } $
  • $\log (x + \log \sec x) + c$
  • B
    $ - \log (x + \log \sec x) + c$
  • C
    $\log (x - \log \sec x) + c$
  • D
    None of these

Answer

Correct option: A.
$\log (x + \log \sec x) + c$
a
(a) 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.$

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