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Math Help - Help with some integrals

  1. #1
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    Help with some integrals

    Well I can't get the math latex to do what I want so here it is in laymans terms.

    I am trying to help a friend with some what seems like basic calc stuff only it's been a while since I've done calc 1 and 2...

    So I have several questions which I'll break down into three different new topics..

    I am trying to evaluate the indefinite integral of 2X/X+3 dx

    I put X+3 in the numerator like so (X+1)^ -1 but I'm not sure what to do from here... or if thats right.


    Second question:

    The definite integral evaluated from 7 to 10 of the absolute value of (X^2 - 81)/(X+9) dx

    I realize that factors but would you do that? It would leave you with ((X+9)(X-9))/(X+9) which ultimate just leaves one factor of X-9 in the numerator.

    So it would be the definite integral evaluated from 7 to 10 of absolute val (X-9)...



    Any help would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by battleman13 View Post
    Well I can't get the math latex to do what I want so here it is in laymans terms.

    I am trying to help a friend with some what seems like basic calc stuff only it's been a while since I've done calc 1 and 2...

    So I have several questions which I'll break down into three different new topics..

    I am trying to evaluate the indefinite integral of 2X/X+3 dx

    I put X+3 in the numerator like so (X+1)^ -1 but I'm not sure what to do from here... or if thats right.


    Second question:

    The definite integral evaluated from 7 to 10 of the absolute value of (X^2 - 81)/(X+9) dx

    I realize that factors but would you do that? It would leave you with ((X+9)(X-9))/(X+9) which ultimate just leaves one factor of X-9 in the numerator.

    So it would be the definite integral evaluated from 7 to 10 of absolute val (X-9)...



    Any help would be greatly appreciated!
    For the first one you either need to do long division or algebraic manipulations to get

    \frac{2x}{x+3}=2\frac{x}{x+3}=2\frac{(x+3)-3}{x+3}=2\frac{x+3}{x+3}+2\frac{-3}{x+3}=2-\frac{3}{x+3}

    As I said you can also use polynomial long division.

    You are correct on the 2nd one just be careful about where |x-9| changes signs
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  3. #3
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    Thanks! That helps a lot
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