Question

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

Answer 1

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

$=\int (e^{\left(\log a^x\right)}+e^{\left(\log x^a\right)}+e^{\left(\log a^a\right)})dx[\because \log x^a=a\log x]$

$=\int (a^x+x^a+a^a)dx[\because e^{\log x}=x]$

$=\int a^{x}dx+\int x^{a}dx+\int a^{a}dx[\because \int (f\left(x\right)+g\left(x\right))dx=\int f\left(x\right)dx+\int g\left(x\right)dx]$

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

Hence $\int (e^{x\log a}+e^{a\log x}+e^{a\log a})dx=\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+a^{a}x+C$