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Math Help - Integration by Partial Fractions

  1. #1
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    Integration by Partial Fractions

    \int \frac{x^3}{(x+1)^{10}} dx

    The only way i can see to solve this would be to use partial fractions, but that would take a really long time. Is there another way?
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  2. #2
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    Quote Originally Posted by twoteenine View Post
    \int \frac{x^3}{(x+1)^{10}} dx

    The only way i can see to solve this would be to use partial fractions, but that would take a really long time. Is there another way?
    Use the substitution: u=x+1
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  3. #3
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    I tried that but then I got that du=dx and then I had...

    \int \frac{x^3}{(u)^{10}} du<br />

    How do you change the x^3 to terms of u?

    Edit: Nevermind i've got it, x = u-1. That was kind of obvious haha, thanks.
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  4. #4
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    Hello, twoteenine!

    Just for fun, I integrated it by parts.

    . . I must get a life!

    I \;=\;\int \frac{x^3}{(x+1)^{10}}\,dx

    . . \begin{array}{ccccccc}u &=& x^3 && dv &=& (x+1)^{10}dx \\ du &=& 3x^2\,dx && v &=& \text{-}\frac{1}{9}(x+1)^{-9} \end{array}


    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\underbrace{\int x^2(x+1)^{-9}\,dx}_{\text{Integrate by parts}}

    . . \begin{array}{ccccccc}u &=& x^2 && dv &=& (x+1)^{-9}\,dx \\ du &=& 2x\,dx && v &=&\text{-}\frac{1}{8}(x+1)^{-8} \end{array}


    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\bigg[-\frac{1}{8}x^2(x+1)^{-8} + \frac{1}{4}\int x(x+1)^{-8}\,dx\bigg]

    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\underbrace{\int x(x+1)^{-8}\,dx}_{\text{Integrate by parts}}

    . . \begin{array}{ccccccc}u &=& x && dv &=& (x+1)^{-8}\,dx \\ du &=& dx && v &=& \text{-}\frac{1}{7}(x+1)^{-7} \end{array}


    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\bigg[-\frac{1}{7}x(x+1)^{-7} + \frac{1}{7}\int (x+1)^{-7}\,dx\bigg]

    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} + \frac{1}{84}\int(x+1)^{-7}\,dx

    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} - \frac{1}{504}(x+1)^{-6} + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Technically, we're done,
    . . but most textbooks simplify the answer beyond all recognition.


    Factor:

    . . I \;=\;-\frac{1}{504}(x+1)^{-9}\bigg[56x^3 + 21x^2(x+1) + 6x(x+1)^2 + (x+1)^3\bigg] + C

    . . . . =\; -\frac{1}{504}(x+1)^{-9}\left[84x^3 + 36x^2 + 9x + 1\right] + C

    . . . . =\;-\,\frac{84x^3 + 36x^2 + 9x + 1}{504(x+1)^9} + C

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, twoteenine!

    Just for fun, I integrated it by parts.
    . . I must get a life!


    . . \begin{array}{ccccccc}u &=& x^3 && dv &=& (x+1)^{10}dx \\ du &=& 3x^2\,dx && v &=& \text{-}\frac{1}{9}(x+1)^{-9} \end{array}


    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\underbrace{\int x^2(x+1)^{-9}\,dx}_{\text{Integrate by parts}}

    . . \begin{array}{ccccccc}u &=& x^2 && dv &=& (x+1)^{-9}\,dx \\ du &=& 2x\,dx && v &=&\text{-}\frac{1}{8}(x+1)^{-8} \end{array}


    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} + \frac{1}{3}\bigg[-\frac{1}{8}x^2(x+1)^{-8} + \frac{1}{4}\int x(x+1)^{-8}\,dx\bigg]

    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\underbrace{\int x(x+1)^{-8}\,dx}_{\text{Integrate by parts}}

    . . \begin{array}{ccccccc}u &=& x && dv &=& (x+1)^{-8}\,dx \\ du &=& dx && v &=& \text{-}\frac{1}{7}(x+1)^{-7} \end{array}


    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} + \frac{1}{12}\bigg[-\frac{1}{7}x(x+1)^{-7} + \frac{1}{7}\int (x+1)^{-7}\,dx\bigg]

    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} + \frac{1}{84}\int(x+1)^{-7}\,dx

    I \;=\;-\frac{1}{9}x^3(x+1)^{-9} - \frac{1}{24}x^2(x+1)^{-8} - \frac{1}{84}x(x+1)^{-7} - \frac{1}{504}(x+1)^{-6} + C


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Technically, we're done,
    . . but most textbooks simplify the answer beyond all recognition.


    Factor:

    . . I \;=\;-\frac{1}{504}(x+1)^{-9}\bigg[56x^3 + 21x^2(x+1) + 6x(x+1)^2 + (x+1)^3\bigg] + C

    . . . . =\; -\frac{1}{504}(x+1)^{-9}\left[84x^3 + 36x^2 + 9x + 1\right] + C

    . . . . =\;-\,\frac{84x^3 + 36x^2 + 9x + 1}{504(x+1)^9} + C

    Your first v is wrong.

    Edit: I mean the dv.
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