Question:
Write a value of $\int\left(e^{x \log _{4} a}+e^{a \log _{4} x}\right) d x$
Solution:
We know that by using property of logarithm
$e^{x \log _{e} a}=e^{\log _{e} a^{x}}=a^{x}$ and $e^{a \log _{e} x}=e^{\log _{e} x^{a}}=x^{a}$
$y=\int a^{x}+x^{a} d x$
$y=\int a^{x} d x+\int x^{a} d x$
Use formula $\int a^{x} d x=\frac{a^{x}}{\log a}$ and $\int x^{a} d x=\frac{x^{a+1}}{a+1}$
$y=\frac{a^{x}}{\log a}+\frac{x^{a+1}}{a+1}+c$
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