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• 1 Post By MarkFL
• 1 Post By HallsofIvy
• 1 Post By Soroban

# Thread:

1. ## 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

2. ## 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

3. ## Re: When integreating by parts, how to choose u and v'?

oh thank you , I have been guessing all this time.

4. ## 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

5. ## Re: When integreating by parts, how to choose u and v'?

We are given:

$\int xe^x\,dx$

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

So I would choose:

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

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

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

It appears in the image you attached, that they are using the reverse of the variables I was taught.

6. ## 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 $\int u dv= uv- \int vdu$ you want to choose u that can be differentiated, dv that can be integrated, and then check to see if you can integrate $\int vdu$.

7. ## Re: When integreating by parts, how to choose u and v'?

Hello, anees7112!

I never really understood how to choose the $u$ and the $dv$ 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 $dv$ be the "hardest" part you can integrate.

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

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

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

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

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

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

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

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

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

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

The new integral is worse than the original integral!

8. ## 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 .

9. ## 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
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