MCQ
$\int_{}^{} {\frac{{{e^x}(x + 1)}}{{{{\cos }^2}(x{e^x})}}dx = } $
  • $\tan (x{e^x}) + c$
  • B
    $\sec (x{e^x})\tan (x{e^x}) + c$
  • C
    $ - \tan (x{e^x}) + c$
  • D
    None of these

Answer

Correct option: A.
$\tan (x{e^x}) + c$
a
(a)$\int_{}^{} {\frac{{{e^x}(x + 1)}}{{{{\cos }^2}(x{e^x})}} = \int_{}^{} {{e^x}(x + 1){{\sec }^2}(x{e^x})dx} } $
Putting $x{e^x} = t \Rightarrow (x + 1){e^x}dx = dt$, we get
$\int_{}^{} {{{\sec }^2}t\,dt = \tan t + c} = \tan (x{e^x}) + c.$

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