Integral Calculus Question 140
Question: $ \int_{{}}^{{}}{({e^{a\log x}}+{e^{x\log a}})dx}= $
Options:
A) $ {x^{a+1}}+\frac{a^{x}}{\log a}+c $
B) $ \frac{{x^{a+1}}}{a+1}+a^{x}\log a+c $
C) $ \frac{{x^{a+1}}}{a+1}+\frac{a^{x}}{\log a}+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{({e^{a\log x}}+{e^{x\log a}}),dx}=\int_{{}}^{{}}{({e^{{\log_{e}}x^{a}}}+{e^{{\log_{e}}a^{x}}}),dx} $ $ =\int_{{}}^{{}}{(x^{a}+a^{x}),dx}=\frac{{x^{a+1}}}{a+1}+\frac{a^{x}}{\log a}+c $ .
BETA
