Recent content by Electgineer99

Okay I got it now, u° equal 1, so the answer is 3×.

∫3×ln3.dx u=3× dx=du/ln(3).3× ∫u.ln(3).du/ln(3).3× ln(3) canceled each other. ∫u.du/3× ⅓×∫u.du ⅓×.u²/2 Then I would substitute u=3× Am I right with this?

it is base 10, but will it make any difference if it is base 10 or any other bases?

The equation is 3^xlog3

|3^xlog3dxI don't even know where to start. I know that the formula is |u.dv = uv - |v.du u=3^x v=log3