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- Thread starter Electgineer99
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- #2

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[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.

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- #5

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[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

- #6

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- The equation is 3^xlog3

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

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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?

- #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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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

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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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That integrand is 3 times ln(3) .

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 think you mean for it to be 3

After the u substitution you have that du = 3

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

- #13

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

- #14

SammyS

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

Also, don't forget the constant of integration.

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