Understanding Integration by Parts: Exploring the Formula and Solving Examples

[Electgineer99]

|3^xlog3dxI don't even know where to start.
I know that the formula is
|u.dv = uv - |v.du
u=3^x v=log3

 

[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.

 

[BiGyElLoWhAt]

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

 

[Delta2]

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

 

[Electgineer99]

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

 

[Mastermind01]

1. The equation is 3^xlog3

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

 

[Electgineer99]

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]

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

 

[fresh_42]

$c = \log_{10} 3$ 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 $a = \ln 3.$

 

[Mastermind01]

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.

 

[Electgineer99]

∫3×ln3.dx
u=3× dx=du/ln(3).3×
∫u.ln(3).du/ln(3).3×
ln(3) canceled each other.
∫u.du/3×
⅓×∫u.du
⅓×.u²/2
Then I would substitute u=3×
Am I right with this?

 

[SammyS]

∫3×ln3.dx
u=3× dx=du/ln(3).3×
∫u.ln(3).du/ln(3).3×
ln(3) canceled 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 3^x ln(3) .

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

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

 

[Electgineer99]

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

 

[SammyS]

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

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

Also, don't forget the constant of integration.