Mathematics

# $\int(e^{a \log x}+ e^{x \log a})dx$

##### SOLUTION
$\int (e^{a \log x}+e^{x \log a})dx$
Assume $\log$ on base $e$
$a \log x = ln x^{a}$
$x \log a= ln a^{x}$
$=\int (e^{lnx^{a}}+e^{lna^{x}})dx$
we know that $e^{ln(f(x))}=f(x)$
$=\int (x^{a}+a^{x})dx$
$=\dfrac{x^{a+1}}{a+1}+\dfrac{a^{x}}{ln \ a}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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