Integration by parts, why is this allowed?

find_the_fun

Active member
I'm following this example where it is asked to integrate $$\displaystyle \int \ln{x} dx$$ 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

Staff member
Re: integration by parts, why is this allowed?

We are actually setting:

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

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

find_the_fun

Active member
Re: integration by parts, why is this allowed?

We are actually setting:

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

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

Staff member
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:

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

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

HallsofIvy

Well-known member
MHB Math Helper
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

Well-known member
MHB Math Helper
If it helps you can think of

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

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