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Thread: Need Help With Indefinite Integral

  1. #1
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    Need Help With Indefinite Integral

    This particular integral is driving me crazy. I can't figure it out for the life of me. Could someone please explain to me how to go about solving it?

    Integrate[5x(7x - 9)^14dx

    I was given a few hints but they still are not helping me at all. First let U=7x-9 and then let x=(U+9)/7. I would imagine this should be done using the substitution method. Thanks in advance

    John
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  2. #2
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    Quote Originally Posted by TreeMoney
    This particular integral is driving me crazy. I can't figure it out for the life of me. Could someone please explain to me how to go about solving it?

    Integrate[5x(7x - 9)^14dx

    I was given a few hints but they still are not helping me at all. First let U=7x-9 and then let x=(U+9)/7. I would imagine this should be done using the substitution method. Thanks in advance

    John
    Is that
    INT.[5x (7x -9)^14]dx ?

    If yes, then here is one way of doing it.

    Integration by parts:
    INT.[u]dv = u*v -INT.[v]du ------***

    Let u = 5x
    So, du = 5dx

    And dv = (7x -9)^14 dx = (7x-9)^14 *(7dx /7) = (1/7)(7x-9)^14 *(7dx)
    So, v = (1/7)[(1/15)(7x-9)^15] = (1/105)(7x-9)^15

    Hence,
    INT.[5x (7x-9)^14]dx
    = (5x)(1/105)(7x-9)^15 -INT.[(1/105)(7x-9)^15](5dx)
    = (1/21)x(7x-9)^15 -(1/21)INT[(7x-9)^15]dx
    = (1/21)x(7x-9)^15 -(1/21){(1/7)INT[(7x-9)^15](7dx)}
    = (1/21)x(7x-9)^15 -(1/147){(1/16)(7x-9)^16}
    = (1/21)x(7x-9)^15 -(1/2352)(7x-9)^16 -------------answer.

    Check,
    d/dx {(1/21)x(7x-9)^15 -(1/2352)(7x-9)^16}
    = (1/21){[x*15(7x-9)^14 *7] +[(7x-9)^15 *1]} -(1/2352){16(7x-9)^15 *7}
    = {5x(7x-9)^14 +(1/21)(7x-9)^15} -(1/21)(7x-9)^15
    = 5x(7x-9)^14
    Okay.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TreeMoney
    This particular integral is driving me crazy. I can't figure it out for the life of me. Could someone please explain to me how to go about solving it?

    Integrate[5x(7x - 9)^14dx

    I was given a few hints but they still are not helping me at all. First let U=7x-9 and then let x=(U+9)/7. I would imagine this should be done using the substitution method. Thanks in advance

    John
    It can be done by integration by parts, but also directly by substitution.
    $\displaystyle \int dx \, 5x(7x-9)^{14} = \int \left ( \frac{du}{7} \right ) \, 5 \left ( \frac{u+9}{7} \right ) u^{14}$
    $\displaystyle =\frac{5}{7} \int du \, \left ( \frac{u}{7} + \frac{9}{7} \right )u^{14}$
    $\displaystyle =\frac{5}{7} \int du \, \frac{1}{7}u^{15} + \frac{5}{7} \int du \, \frac{9}{7}u^{14}$
    $\displaystyle =\frac{5}{7} \frac{1}{7}\frac{1}{16}u^{16} + \frac{5}{7} \frac{9}{7}\frac{1}{15}u^{15}$
    $\displaystyle =\frac{5}{784}(7x-9)^{16}+\frac{3}{49}(7x-9)^{15}$

    Which you can then simplify any way you wish.

    -Dan
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  4. #4
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    Thanks Guys. Now that I see how it is done, I was making stupid algebra mistakes. Thanks again for the help. It's much appreciated.

    John
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TreeMoney
    Thanks Guys. Now that I see how it is done, I was making stupid algebra mistakes. Thanks again for the help. It's much appreciated.

    John
    There's no such thing as a "stupid mistake." You either get it right or you don't. You didn't get it right, you knew you didn't, and you asked a question about it. I see no stupidity there, I see a good practice in action!

    -Dan
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