ज्ञात करें `int(e^("x log a")+e^("a log x"))dx,ane1`

Explanation

`int(e^("x log a"))+e^("x log a")dx=inte^("x log a")+ int ^(log(x^a)dx)`
`=e^(" x log a")/("log a") + int x^adx=a^x/("log a")+x^(a+1)/(a+1)+c`