Welcome to our community

Be a part of something great, join today!

Integration by parts, why is this allowed?

find_the_fun

Active member
Feb 1, 2012
166
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Feb 1, 2012
166
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 29, 2012
1,151
Any time you use "integration by parts",
[tex]\int udv= uv- \int v du[/tex], 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
Mar 1, 2012
249
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 \(\displaystyle u = \ln(x)\)