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Math Help - Help on an indefinte integral

  1. #1
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    Help on an indefinte integral

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

    Can someone help me with a starting point on this one please? Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle \frac{x^{2} + 2 x}{(x+1)^{2}} = 1-\frac{1}{(x+1)^{2}}

    ... so that...

    Kind regards

    \chi \sigma
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  3. #3
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    Can you show me how those are equal?
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  4. #4
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    Quote Originally Posted by dbakeg00 View Post
    Can you show me how those are equal?
    You should be able to do the algebra at this level.


    \displaystyle \frac{x^2 + 2x}{x^2 + 2x + 1} = \frac{(x^2 + 2x + 1) - 1}{x^2 + 2x + 1} = ....
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Is...

    \displaystyle \frac{x^{2} + 2 x}{(x+1)^{2}} = 1-\frac{1}{(x+1)^{2}}

    ... so that...

    Kind regards

    \chi \sigma
    \displaystyle\int 1-\frac{1}{(x+1)^2}dx

    \displaystyle\int 1\ dx - \int (x+1)^{-2}dx

    \displaystyle x+(x+1)^{-1}+c
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Well done
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  7. #7
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    The book is giving me an answer of:

    \displaystyle\frac{x^2}{(x+1)}+c
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  8. #8
    MHF Contributor Unknown008's Avatar
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    \begin{array}{cl}x+(x+1)^{-1}+c &= x + \dfrac{1}{x+1} + c \\<br /> <br />
& \\<br /> <br />
&= \dfrac{x(x+1) + 1}{x+1} + c \\ & \\<br /> <br />
& = \dfrac{x^2 + x + 1}{x+1} + c \end{array}

    There must be a mistake in the book.

    EDIT: It can be broken into two fractions:

     \dfrac{x^2 + x + 1}{x+1} + c  = \dfrac{x^2}{x+1} + \dfrac{x+1}{x+1} + c = \dfrac{x^2}{x+1} +1 + c = \dfrac{x^2}{x+1} +c'
    Last edited by Unknown008; November 16th 2010 at 02:44 AM.
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  9. #9
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    Ok. I came up with the same thing when I expanded it. Just wanted to make sure I wasn't overlooking something. Thanks again for the help!
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  10. #10
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    Quote Originally Posted by dbakeg00 View Post
    The book is giving me an answer of:

    \displaystyle\frac{x^2}{(x+1)}+c
    \displaystyle \frac{x^2}{(x+1)}+c is equivalent to \displaystyle x + \frac{1}{(x+1)} + d:

    \displaystyle x + \frac{1}{(x+1)} + d = \frac{x^2 + x + 1}{x + 1} + d = \frac{x^2 + (x + 1)}{x + 1} + d = \frac{x^2}{x + 1} + 1 + d = \frac{x^2}{x + 1} + c.
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