Integral Calculus Question 122
Question: $ \int_{{}}^{{}}{\frac{{x^{e-1}}+{e^{x-1}}}{x^{e}+e^{x}}dx=} $
Options:
A) $ \log (x^{e}+e^{x})+c $
B) $ e\log (x^{e}+e^{x})+c $
C) $ \frac{1}{e}\log (x^{e}+e^{x})+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
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 $ .
BETA
