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You don't need by parts to do this. You can use a simple substitution.
[tex]\int 3^x ln 3 dx[/tex]
Let [itex]u = 3^x[/itex] , then [itex] du = 3^x ln 3 dx[/itex]
The integral becomes [tex]\int du [/tex] Integrate and substitute u back.
- The equation is 3^xlog3
it is base 10, but will it make any difference if it is base 10 or any other bases?That's what I wrote, or do you mean to say it's base 10?
it is base 10, but will it make any difference if it is base 10 or any other bases?
That integrand is 3 times ln(3) .∫3×ln3.dx
u=3× dx=du/ln(3).3×
∫u.ln(3).du/ln(3).3×
ln(3) cancelled each other.
∫u.du/3×
⅓×∫u.du
⅓×.u²/2
Then I would substitute u=3×
Am I right with this?
I believe that you mean 3x or write 3^x .Okay I got it now, u° equal 1, so the answer is 3×.