Question
Write a value of $\int(\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}})\text{dx}$
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Answer
Let $\text{I}=\int\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}}\text{ dx}$
$=\int(\text{e}\log\text{a}^{\text{x}}+\text{e}\log\text{x}^{\text{a}})\text{dx}$
$=\int(\text{a}^{\text{x}}+\text{x}^{\text{a}})\text{dx}$
$=\int\text{a}^{\text{x}}\text{dx}+\int\text{x}^{\text{a}}\text{dx}$
$=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
$\therefore\ \text{I}=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
$=\int(\text{e}\log\text{a}^{\text{x}}+\text{e}\log\text{x}^{\text{a}})\text{dx}$
$=\int(\text{a}^{\text{x}}+\text{x}^{\text{a}})\text{dx}$
$=\int\text{a}^{\text{x}}\text{dx}+\int\text{x}^{\text{a}}\text{dx}$
$=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
$\therefore\ \text{I}=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
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