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

Answer

Correct option: C.
$\frac{1}{e}\log ({x^e} + {e^x}) + c$
c
(c) Put ${x^e} + {e^x} = t \Rightarrow e({x^{e - 1}} + {e^{x - 1}})\,dx = dt,$
$\int_{}^{} {\frac{{{x^{e - 1}} + {e^{x - 1}}}}{{{x^e} + {e^x}}}} \,dx = \frac{1}{e}\int_{}^{} {\frac{{dt}}{t}} = \frac{1}{e}\log t = \frac{1}{e}\log ({x^e} + {e^x}) + c$.

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