If `a >0` and `a!=1` evaluate the following integrals: `inte^x\ a^x\ dx` (ii) `int2^((log)_e x)\ dx`

Explanation

i) `\int e^{x} a^{x} d x \quad\{\int a^{x} d x=\frac{a^{x}}{\log _{e} a^{x}}\}`


`=\int(e a)^{x} d x=\frac{\mid {aet} x}{\log _{e}(a c)}+c`


ii) `\int 2^{\log _{e}^{x}} d x=\frac{x^{\log _{e}^{2}+1}}{\log _{e}^{2}+1}+c`


`\text { to }\{x^{\log _{e} y}=y^{\log _{e} x}\}`