When integreating by parts, how to choose u and v'?

• Dec 7th 2012, 08:30 PM
anees7112
When integreating by parts, how to choose u and v'?
Hello, I never really understood how to choose the u and the v' when integrating by parts. I know if you choose the wrong one sometimes you end up in a loop.

For the question

Integrate ln x dx
It becomes : Integrate 1 * ln x dx

Now do I set u' as 1 or as ln x? And how do you decide which one is which?

Thank you :)
• Dec 7th 2012, 08:48 PM
MarkFL
Re: When integreating by parts, how to choose u and v'?
A general rule for which function to choose as u is given by the mnemonic LIATE:

Integration by parts - Wikipedia, the free encyclopedia
• Dec 7th 2012, 08:53 PM
anees7112
Re: When integreating by parts, how to choose u and v'?
oh thank you , I have been guessing all this time.
• Dec 7th 2012, 11:26 PM
anees7112
Re: When integreating by parts, how to choose u and v'?
Sorry one more thing.

How come in this question they set u' as the exponential rather than the algebraic Attachment 26135
• Dec 8th 2012, 12:34 AM
MarkFL
Re: When integreating by parts, how to choose u and v'?
We are given:

$\displaystyle \int xe^x\,dx$

I was taught: $\displaystyle \int u\,dv=uv-\int v\,du$

So I would choose:

$\displaystyle u=x\,\therefore\,du=dx$

$\displaystyle dv=e^x\,dx\,\therefore\,v=e^x$ giving:

$\displaystyle \int xe^x\,dx=xe^x-\int e^x\,dx=e^x(x-1)+C$

It appears in the image you attached, that they are using the reverse of the variables I was taught.
• Dec 8th 2012, 05:06 AM
HallsofIvy
Re: When integreating by parts, how to choose u and v'?
"Guessing" is not a bad method, as long as you check! In general, because $\displaystyle \int u dv= uv- \int vdu$ you want to choose u that can be differentiated, dv that can be integrated, and then check to see if you can integrate $\displaystyle \int vdu$.
• Dec 8th 2012, 07:25 AM
Soroban
Re: When integreating by parts, how to choose u and v'?
Hello, anees7112!

Quote:

I never really understood how to choose the $\displaystyle u$ and the $\displaystyle dv$ when integrating by parts.
I know if you choose the wrong one sometimes you end up in a loop.

Here's the rule that I was taught:

Let $\displaystyle dv$ be the "hardest" part you can integrate.

Example: .$\displaystyle I \:= \int\!\! x\,e^{3x}dx$

We can integrate $\displaystyle x$ and we can integrate $\displaystyle e^{3x}.$
. . The harder one is $\displaystyle e^{3x}.$

We have: .$\displaystyle \begin{Bmatrix} u &=& x && dv &=& e^{3x}dx \\ du &=& dv && v &=& \frac{1}{3}e^{3x}\end{Bmatrix}$

Then: .$\displaystyle I \;=\;\tfrac{1}{3}xe^{3x} - \tfrac{1}{3}\!\int\! e^{3x}dx$

. . . . . . . $\displaystyle =\;\tfrac{1}{3}xe^{3x} - \tfrac{1}{9}e^{3x} + C \;=\;\tfrac{1}{9}e^{3x}(3x-1) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Same example: .$\displaystyle I \;=\;\int x\,e^{3x}\,dx$

Watch what happens if you do it "the other way" . . .

You have: .$\displaystyle \begin{Bmatrix}u &=& e^{3x} && dv &=& x\,dx \\ du &=& 3e^{3x}dx && v &=& \frac{1}{2}x^2\end{Bmatrix}$

$\displaystyle \text{Then: }\:I \;=\;\tfrac{1}{2}x^2e^{3x} - \tfrac{3}{2}\!\underbrace{\int\!x^2e^{3x}\,dx}_{?? }$

The new integral is worse than the original integral!
• Dec 10th 2012, 06:24 AM
anees7112
Re: When integreating by parts, how to choose u and v'?
Thank you all for the help :) I probably should have asked this question a few years ago when we learnt integration by parts but it feels good to finally get it .
• Dec 10th 2012, 06:20 PM
hollywood
Re: When integreating by parts, how to choose u and v'?
I have always just "guessed", or more accurately, set u and dv to what seemed easy to differentiate and integrate. It wasn't until I actually taught Calculus that I learned about the LIATE acronym. One of my students told me about it.

- Hollywood