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Homework Help: Integration by parts

  1. May 4, 2009 #1
    Formula for integration by parts:

    [tex]\int f(x)dx = \int u dv = uv - \int v du[/tex]

    Use integration by parts to find the following integrals:

    a) [tex]\int x e^{1-x} dx [/tex]

    b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

    c) [tex]\int_{-2}^1 (2x+1)(x+3)^{3/2} dx[/tex]

    d) [tex]\int x^3 \sqrt{3x^2+2} dx[/tex]

    Answers in back of the book:

    a) [tex]-e^{1-x}(x+1)+C[/tex]

    b) 4 ln 2 - 2

    c) [tex]\frac{74}{4}[/tex]

    d) [tex]\frac {1}{9}x^2(3x^2+2)^{\frac {3}{2}}-\frac{2}{135}(3x^2+2)^{\frac {5}{2}} +C[/tex]

    My attempts:

    a) [tex]\int xe^{(1-x)} dx[/tex]

    [tex]u = x[/tex]

    [tex]dv = e^{1-x} dx[/tex]

    [tex]du = 1 [/tex]

    [tex]v = -e^{1-x}[/tex]

    which gives:

    [tex]-xe^{1-x} - \int -e^{1-x}(1)[/tex]

    [tex]-xe^{1-x} - e^{1-x}[/tex]

    [tex]-e^{1-x}(x+1) +C[/tex]

    GOT IT! Thanks Melawrghk and dx!

    b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

    [tex]u = ln x^{{1}{2}}[/tex]

    [tex]dv = x^{\frac {-1}{2}}[/tex]

    [tex]du = \frac { \frac {1}{2}\sqrt x}{\sqrt x} = \frac {1}{2x}[/tex]

    [tex]v = 2x^{\frac{1}{2}}[/tex]

    which gives:

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int 2x^{\frac{1}{2}} \frac {1}{2x}[/tex]

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int \frac{x^{\frac {1}{2}}}{x}[/tex]

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int x^{\frac {-1}{2}}[/tex]

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - 2x^{\frac{1}{2}}[/tex]

    [tex] 2x^{\frac{1}{2}}(ln x^{\frac {1}{2}}-1)[/tex]

    Plug in 4 and 1:


    [tex] = 4ln2-4+2[/tex]

    [tex] = 4ln2-2[/tex]

    GOT IT! Thanks again dx!!

    I'm looking at the other two and I'm blanking.

    I've been out of school for three years, and jumped right into business calculus since it's a requirement. So I apologize if this is elementary, but I've been banging my head on this for hours now... and of course the teacher is nowhere to be found. :cry:

    Thanks in advance!

    EDIT: Got a correct now, thanks to Melawrghk and dx.
    Last edited: May 4, 2009
  2. jcsd
  3. May 4, 2009 #2
    For a), switch your u and dv and try again :)

    EDIT: and for b), check your derivative of ln(sqrt(x)). sqrt(x) can be written as x^(1/2), so when you take the derivative you'll get (1/2)x^(3/2). Coupled with the square root of x on the bottom it will cancel nicely.
  4. May 4, 2009 #3


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    For part (a), v = e1-x would be a better choice.
  5. May 4, 2009 #4
    Is [tex] \int e^{1-x} = e^{1-x}[/tex] ?

    Edit: I'm only assuming this because

    [tex] \int e^{x} = e^{x}[/tex]

    ... but now that I think about it...

    [tex] \int e^{1-x} = \frac {1}{1-x} e^{1-x+1} = \frac {1}{1-x} e^{x}[/tex]

    Last edited: May 4, 2009
  6. May 4, 2009 #5
    Wouldn't the derivative of x^(1/2) = (1/2)x^(-1/2)?
  7. May 4, 2009 #6


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    Nope. ∫ e1-x dx = -e1-x. Use the substitution t = 1-x to see why.
  8. May 4, 2009 #7


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    Yes, that's correct.
  9. May 4, 2009 #8
    Aha! Great, now I got part a and edited up there. Thanks :)
  10. May 4, 2009 #9


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    For part (b), use v = √x.
  11. May 4, 2009 #10
    Could you walk me through this method, please?

    I'm taught to look at the equation and assign u and dv. Then derive u to get du and integrate dv to get v.

    The book says:

    Step 1: Choose functions u and v so that f(x)dx = u dv. Try to pick u so that du is simpler than u and a dv that is easy to integrate.

    Step 2: Organize the computation... (they put it in a neat box :tongue:)

    Step 3: Complete the integration by finding ∫f(x) dx = ∫ u dv = uv - ∫ v du

    So I'm not sure how I assign v = √x directly.

    Here's the problem again:

    \int_1^4 \frac {ln \sqrt x} {\sqrt x} dx
  12. May 4, 2009 #11


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    Hint: If v = √x, then

    [tex] dv = \frac{dx}{2\sqrt{x}}[/tex]
  13. May 4, 2009 #12
    Got part b! Dx, you rock!
  14. May 4, 2009 #13
    OK so for part d I'm running into a seemingly basic problem. I'm not sure how to integrate √(3x2+2)

    I'm just fried....
  15. May 4, 2009 #14
    You have a direct formula for that.
  16. May 4, 2009 #15
    What is the formula? I'm trying to study for tomorrow morning's final, doing these practice questions and papers are everywhere.
  17. May 5, 2009 #16
    http://planetmath.org/encyclopedia/IntegrationFormulas.html [Broken]
    Last edited by a moderator: May 4, 2017
  18. May 5, 2009 #17


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    Don't use a formula, splitting it up in x^3 and the root is a poor way to split things. Do it like this instead.

    [tex]x^3 \sqrt{3x^2+2}=x^2*x\sqrt{3x^2+2}[/tex]. Now use partial integration and note that [itex]x\sqrt{3x^2+2}[/itex] is really easy to integrate by substituting u=3x^2+2, or even "guessing" the primitive.
    Last edited: May 5, 2009
  19. May 5, 2009 #18
    Oops.. My apologies... I just read the post no 13 and not the original post. So i just thought its [itex]sqrt{3x^2+2}[/itex] and didnt notice the x cube
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