_ ^ 1 x ^2 (1 + x) ;dx =

MCQ

$\int_{}^{} {\frac{1}{{x{{\cos }^2}(1 + \log x)}}\;dx = } $

✓
$\tan \,(1 + \log x) + c$
B
$\cot \,(1 + \log x) + c$
C
$ - \tan \,(1 + \log x) + c$
D
$ - \cot (\,1 + \log x) + c$

✓

Answer

Correct option: A.

$\tan \,(1 + \log x) + c$

a
(a) 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.$

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