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$I=\int {(x)}'3^{lnx}dx=3^{lnx}x-ln3\int3^{lnx}dx $<=>$I=x3^{lnx}-ln3I$<=>$I=\frac{x3^{lnx}}{1+ln3}+c$
$I=\int {(x)}'3^{lnx}dx=3^{lnx}x-ln3\int3^{lnx}dx $<=>
ΑπάντησηΔιαγραφή$I=x3^{lnx}-ln3I$<=>$I=\frac{x3^{lnx}}{1+ln3}+c$