_ ^ x^ e - 1 + e^ x - 1 x^e + e^x dx = - Vidyadip

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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