MHB Integration by parts, why is this allowed?

[find_the_fun]

I'm following this example where it is asked to integrate [math]\int \ln{x} dx[/math] using integration by parts. I don't understand how it's legal to set v=x since the only x in the equation is the argument of ln and that's already accounted for by u.

 

[MarkFL]

Re: integration by parts, why is this allowed?

We are actually setting:

$$u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$$

$$dv=dx\,\therefore\,v=x$$

 

[find_the_fun]

Re: integration by parts, why is this allowed?

We are actually setting:

$$u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$$

$$dv=dx\,\therefore\,v=x$$

Ok I think I get it. So with integration by parts you choose dv before you calculate v? In other words dv has to be part of the original equation but v does not?

 

[MarkFL]

Re: integration by parts, why is this allowed?

Ok I think I get it. So with integration by parts you choose dv before you calculate v? In other words dv has to be part of the original equation but v does not?

Correct...integration by parts states:

$$\int u\,dv=uv-\int v\,du$$

So, you want to choose a $u$ and a $dv$, and then from these compute $du$ and $v$.

 

[HallsofIvy]

Any time you use "integration by parts",
\int udv= uv- \int v du, there is NO "v" in the original integral. You get it by your choice of "dv".

Here, the choice is dv= dx, u= ln(x).

 

[SuperSonic4]

If it helps you can think of

$$\int \ln(x) dx = \int \left(1 \times \ln(x) \right)dx$$

Then you can set $$\frac{dv}{dx} = 1$$ and $$u = \ln(x)$$