MCQ
$\int_{}^{} {{e^{x\log a}}.\;{e^x}\;dx} $is equal to
  • A
    ${(ae)^x} + c$
  • $\frac{{{{(ae)}^x}}}{{\log (ae)}} + c$
  • C
    $\frac{{{e^x}}}{{1 + \log a}} + c$
  • D
    None of these

Answer

Correct option: B.
$\frac{{{{(ae)}^x}}}{{\log (ae)}} + c$
b
(b)$\int_{}^{} {{e^{x\log a}}{e^x}dx} = \int_{}^{} {{e^{\log {a^x}}}.\,{e^x}dx} = \int_{}^{} {{a^x}{e^x}dx} $
$ = \int_{}^{} {{{(ae)}^x}dx} = \frac{{{{(ae)}^x}}}{{\log (ae)}} + C.$

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