Integration by parts

  • #1
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|3^xlog3dx


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

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  • #2
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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.
 
  • #3
BiGyElLoWhAt
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Log 3 is a constant, which means that dv = 0.
 
  • #4
Delta2
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You don't need integration by parts to find this, its pretty simple, do you know what is the derivative of ##3^x##?
 
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  • #5
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.
  1. The equation is 3^xlog3
 
  • #7
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?
 
  • #8
vela
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Are you sure? It's pretty common in calculus classes to denote the natural logarithm by log.
 
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  • #9
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##c = \log_{10} 3## would only be a change by a constant factor ##c##.
You can write ##3^x \cdot \log_{10} 3 = e^{x \cdot \ln3} \cdot \log_{10}3## and use ##\int c e^{ax} dx = \frac{c}{a} e^{ax} + const.## with ##a = \ln 3.##
 
  • #10
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it is base 10, but will it make any difference if it is base 10 or any other bases?
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.
 
  • #11
∫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?
 
  • #12
SammyS
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∫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?
That integrand is 3 times ln(3) .

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

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

Thus your integral simply becomes ##\ \int du \ .##
 
  • #13
Okay I got it now, u° equal 1, so the answer is 3×.
 
  • #14
SammyS
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Okay I got it now, u° equal 1, so the answer is 3×.
I believe that you mean 3x or write 3^x .

Also, don't forget the constant of integration.
 

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