Question

Find the integrals (primitives):
$\rm \displaystyle\int (e^{3\log x}-3^{x}+e^{x\log 3} + e^{3\log 3}) \ dx$

Answer 1

As we know that

$3 log(x) = log( {x}^{3} ) \\ \\ {e}^{ log(x) } = x$
so use these properties first to simplify the integration

$\rm \displaystyle\int (e^{3\log x}-3^{x}+e^{x\log 3} + e^{3\log 3}) \ dx \\ \\ \int ( {e}^{ log( {x}^{3} )} - {3}^{x} + {e}^{ log( {3}^{x} ) } + {e}^{ log( {3}^{3} ) })dx \\ \\ = \int ({x}^{3} - {3}^{x} + {3}^{x} + {3}^{3} )dx \\ \\ = \int {x}^{3} dx + \int \: 27dx \\ \\ = \frac{ {x}^{4} }{4} + 27x + C$
Hope it helps you.