Integration by parts

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.

 

Log 3 is a constant, which means that dv = 0.

 

You don't need integration by parts to find this, its pretty simple, do you know what is the derivative of ##3^x##?

 

That's what I wrote, or do you mean to say it's base 10?

 

Are you sure? It's pretty common in calculus classes to denote the natural logarithm by log.

 

It will make a slight difference, using the same substitution the integral would be [itex]\log e [/itex] (base 10) which is also easily integrated as it's a constant.

But, like @vela said you should make sure it's base 10.

 

That integrand is 3 times ln(3) .

I think you mean for it to be 3^x ln(3) .

After the u substitution you have that du = 3^x ln(3) dx .

Thus your integral simply becomes ##\ \int du \ .##

 

I believe that you mean 3^x or write 3^x .

Also, don't forget the constant of integration.