Integration by parts

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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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You don't need by parts to do this. You can use a simple substitution.
$$\int 3^x ln 3 dx$$
Let $u = 3^x$ , then $du = 3^x ln 3 dx$
The integral becomes $$\int du$$ Integrate and substitute u back.

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

Delta2
Homework Helper
Gold Member
You don't need integration by parts to find this, its pretty simple, do you know what is the derivative of ##3^x##?

Mastermind01
You don't need by parts to do this. You can use a simple substitution.
$$\int 3^x ln 3 dx$$
Let $u = 3^x$ , then $du = 3^x ln 3 dx$
The integral becomes $$\int du$$ Integrate and substitute u back.
1. The equation is 3^xlog3

1. The equation is 3^xlog3

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

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?

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

SammyS and Mastermind01
fresh_42
Mentor
2021 Award
##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.##

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 $\log e$ (base 10) which is also easily integrated as it's a constant.

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

∫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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
∫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 \ .##

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

SammyS
Staff Emeritus