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Integration by Parts

  1. Dec 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Which one is correct?

    ##\int (3x+2) (2x+1)^{\frac{1}{2}} dx = \frac{1}{3} (3x+2)(2x+1)^{\frac{3}{2}} - \frac{1}{15} (2x+1)^{\frac{5}{2}} + C##

    or

    ##\int (3x+2) (2x+1)^{\frac{1}{2}} dx = \frac{1}{3} (3x+2)(2x+1)^{\frac{3}{2}} - \frac{1}{5} (2x+1)^{\frac{5}{2}} + C##

    ?????

    2. Relevant equations


    3. The attempt at a solution

    ##\int (3x+2) (2x+1)^{\frac{1}{2}} dx##

    Let

    ##u = 3x+2##

    Then

    ##\frac{du}{dx} = 3##

    or

    ##du = 3 \ dx##

    Let

    ##dv = (2x+1)^{\frac{1}{2}} dx##

    Then

    ##v = \frac{1}{3} (2x+1)^{\frac{3}{2}}##

    So

    ##\int (3x+2) (2x+1)^{\frac{1}{2}} dx##

    ##= uv - \int v \ du##

    ##= (3x+2) \frac{1}{3} (2x+1)^{\frac{3}{2}} - \int \frac{1}{3} (2x+1)^{\frac{3}{2}} (3 \ dx)##

    ##= \frac{1}{3} (3x+2) (2x+1)^{\frac{3}{2}} - \int (2x+1)^{\frac{3}{2}} dx##

    ##= \frac{1}{3} (3x+2) (2x+1)^{\frac{3}{2}} - \frac{1}{5} (2x+1)^{\frac{5}{2}} + C##
     
    Last edited: Dec 18, 2014
  2. jcsd
  3. Dec 18, 2014 #2

    Tom Mattson

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    You'll have an easier time with this is you don't integrate by parts. Do a u-substitution instead with u=2x+1. Much easier.

    But I worked through your steps I and I got what you got.
     
    Last edited: Dec 18, 2014
  4. Dec 20, 2014 #3
    What integration method I should use?

    Do you mean changing the ##u = 2x + 1## instead of ##u = 3x + 2##?

    If so, let me re-work this integration, by part.

    ##\int (3x + 2)(2x + 1)^{\frac{1}{2}}dx##

    Let

    ##u = 2x + 1##

    Then

    ##\frac{du}{dx} = 2##

    Or

    ##du = 2 \ dx##

    ####

    Let

    ##dv = (3x + 2) \ dx##

    Then

    ##v = \frac{3}{2}x^2 + 2x##

    So

    ##\int (3x + 2)(2x + 1)^{\frac{1}{2}}dx##

    ##= uv - \int v \ du##

    ##= (2x + 1) (\frac{3}{2}x^2 + 2x) - \int (\frac{3}{2}x^2 + 2x)(2 \ dx)##

    ##= (2x + 1) (\frac{3}{2}x^2 + 2x) - \int (3x^2 + 4x) dx##

    ##= (2x + 1) (\frac{3}{2}x^2 + 2x) - x^3 - 2x^2 + C##

    ##= x^3 + 4x^2 + \frac{3}{2}x^2 + 2x - x^3 - 2x^2 + C##

    ##= \frac{7}{2}x^2 + 2x + C##

    Is it correct?

    Because this is an integration by part, why use ##u = 2x + 1## instead of ##u = (2x + 1)^{\frac{1}{2}}##?

    Will the result be the same or different?
     
    Last edited: Dec 20, 2014
  5. Dec 20, 2014 #4

    Mark44

    Staff: Mentor

    No, it's not even close. This is something that you can check for yourself. If you differentiate (7/2)x2 + 2x + C, do you get (3x + 2)(2x + 1)1/2?

    When you're doing integration by parts, whatever you choose for u and dv has to multiply to give you the integrand you start with.
    You don't. I think that what Tom Mattson had in mind was an ordinary substitution.
    The answer to this should be obvious.
     
  6. Dec 20, 2014 #5

    Tom Mattson

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    I was just thinking u-substitution. u=2x+1, so x=(u-1)/2 and dx=du/2. Seems like less of a hassle to me because you need 2 u-substitutions in your integration by parts anyway.
     
  7. Dec 27, 2014 #6
    ∫(3x+2)(2x+1)^0.5dx
    let f'(x) = (2x + 1)^0.5 ==> f(x) = ((2x + 1)^1.5)/ 3
    g(x) = 2x + 2 ==> g'(x) = 2
    = (2x + 2)((2x + 1)^1.5) / 3 - 2/3∫(2x + 1)^1.5 dx
    = (2x + 2)((2x + 1)^1.5) / 3 - 1/3(2x + 1)^2.5 + c
    = 1/3((2x + 2)(2x+1)^1.5 - (2x + 1)^ 2.5) + c
     
  8. Jan 2, 2015 #7
    g(x) is not (2x + 2) but (3x + 2).
     
  9. Jan 2, 2015 #8

    Ray Vickson

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    You can easily check for yourself which (if any) of the two answers is correct. Just differentiate both of them and see which (if any) gives you back the original integrand ##(3x+2)(2x+1)^{1/2}##. You should develop the habit of always doing this automatically whenever you do indefinite integrations.
     
  10. Jan 2, 2015 #9

    Mark44

    Staff: Mentor

    I strongly agree.
     
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