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Integration By Parts

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Integrate by parts.


    2. Relevant equations
    (integral) (x^5 * sqrt(x^3 + 5) dx)


    3. The attempt at a solution
    i've tried using simple substitution, not by parts.

    integral (x^3 * x^2 * sqrt(x^3 + 5) dx
    u=x^3 + 5
    du=3x^2

    1/3(integral) (u-5) * u^1/2 du
    1/3(u^3/2 - 5u^1/2)
    --------------------------------------------

    2 (x^3 + 5)^5/2
    --
    15

    (subtract)

    10 (x^3 + 5)^3/2
    --
    3
     
    Last edited: Feb 20, 2013
  2. jcsd
  3. Feb 20, 2013 #2

    SammyS

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    How did you get the (u-5) ?
     
  4. Feb 20, 2013 #3

    Mark44

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    What's your question?

    You can always check your answer to an integration problem by differentiating your answer. If this results in the integrand, then your answer is correct.
     
  5. Feb 20, 2013 #4
    u=x^3 + 5 --> u-5=x^3

    to substitute for the x^3 in the equation i rearrange the above by subtracting 5 from both sides.
     
  6. Feb 20, 2013 #5

    Zondrina

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    u = x^3 + 5, so (1/3)du = x^2dx.

    Notice you can write x^3 = u-5?

    I'm getting : ##\frac{2}{15}(x^3+5)^{5/2} - \frac{10}{9}(x^3+5)^{3/2} + c##
     
  7. Feb 20, 2013 #6
    they have as the answer: (2/9)x^3 * (x^3 + 5)^(3/2) - (4/45)(x^3 + 5)^(5/2) + C.
     
  8. Feb 20, 2013 #7

    Dick

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    Their answer is right and Zondrina's answer is also right. Your's isn't right. There's more than one way of writing an expression. Differentiate your answer to see if it's correct.
     
  9. Feb 20, 2013 #8
    how are those two expressions the same?
     
  10. Feb 20, 2013 #9

    Dick

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    Factor (x^3+5)^(3/2) out of both expressions and rearrange what's left.
     
  11. Feb 21, 2013 #10
    can you please show how?
     
  12. Feb 21, 2013 #11
  13. Feb 21, 2013 #12

    Dick

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    Again, differentiate your answer and see what you get. Then realize you didn't multiply your second term by 1/3. The coefficient should be -10/9, not -10/3. Your result is almost correct, it's just a simple mistake.
     
  14. Feb 21, 2013 #13
    how would i solve this using integration by parts as opposed to u-sub like the problem states.
     
  15. Feb 21, 2013 #14
    I just did it and you can make it work by doing u=x^3 and then you end up with
    [tex]\int u\sqrt{u+5}du[/tex]
    So then you do an integration by parts and it works out.
     
  16. Feb 21, 2013 #15

    SammyS

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    There are two reasonable ways to assign pair of functions, u(x) and v(x) so that [itex]\displaystyle \ \ u(x)v'(x)=x^5\sqrt{x^3+5}\ .[/itex]

    One such assignment is
    [itex]\displaystyle \ \ u(x)=\sqrt{x^3+5}\,,\ [/itex]
    and
    [itex]\displaystyle \ \ v'(x)=x^5\ .[/itex]

    The resulting integral has an integrand of [itex]\displaystyle \ \ v(x)u'(x)\,, \ [/itex] which has x raised to a higher power than the original and a radical in the denominator.

    This is probably not a good choice.​

    The other assignment can be most easily obtained by first assigning v(x), then finding v'(x), with u(x) being what's "left over".
    Let [itex]\displaystyle \ \ v(x)=(x^3+5)^{3/2}\,,\ [/itex] so that v'(x) includes [itex]\displaystyle \ \ \sqrt{x^3+5}\ .[/itex]

    This gives [itex]\displaystyle \ \ v'(x)=\frac{9}{2}x^2\sqrt{x^3+5}\ .\ [/itex]

    That leaves you with [itex]\displaystyle \ \ u(x)=\frac{2}{9}x^3\ .[/itex]​
    See where this second assignment for u(x) and v(x) leads you.
     
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