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

  1. #1
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    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
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  2. #2
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    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
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  3. #3
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    Re: When integreating by parts, how to choose u and v'?

    oh thank you , I have been guessing all this time.
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    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 When integreating by parts, how to choose u and v'?-1231231.png
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  5. #5
    MHF Contributor MarkFL's Avatar
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    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.
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  6. #6
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    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 $\displaystyle \int vdu$.
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  7. #7
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    Re: When integreating by parts, how to choose u and v'?

    Hello, anees7112!

    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!
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  8. #8
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    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 .
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  9. #9
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    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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