∫5×(2/3)^xdx
=5∫e^[ln(2/3)^x]dx
=5∫e^[xln(2/3)]dx
=[5/ln(2/3)]∫{e^[xln(2/3)]}d[xln(2/3)]
=[5/ln(2/3)]e^[xln(2/3)]
=[5/ln(2/3)]×[(2/3)^x]