Question

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

Answer 1

We have,

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

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

$=\int (x^a+a^e)dx$

$=\int x^{a}dx+\int a^{e}dx$

$=\frac{x^{a+1}}{a+1}+a^{e}x+C$

Hence, this is the answer.