Question

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

Answer 1

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

Assume $\log$ on base $e$

$a\log x=ln{x}^a$

$x\log a=ln{a}^x$

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

we know that $e^{ln\left(f\right(x\left)\right)}=f\left(x\right)$

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

$=\frac{x^{a+1}}{a+1}+\frac{a^x}{lna}+c$