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Math Help - integral

  1. #1
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    integral

    evalute:  \int \frac{x + 3}{x^2 + 6 x}dx
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    evalute:  \int \frac{x + 3}{x^2 + 6 x}dx
    use substitution. u = x^2 + 6x
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by viet View Post
    evalute:  \int \frac{x + 3}{x^2 + 6 x}dx
    Or you can use partial fractions:
    x^2 + 6x = x(x + 6)

    So let's find an A and B such that
    \frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}

    Well...
    \frac{A}{x} + \frac{B}{x + 6} = \frac{A(x + 6) + Bx}{x^2 + 6x} = \frac{(A + B)x + 6A}{x^2 + 6x}

    Comparing the two expressions gives me
    A + B = 1
    6A = 3

    Giving me that A = B = \frac{1}{2}.

    So
    \int dx \frac{x + 3}{x^2 + 6x} = \int dx \frac{1/2}{x} + \int dx \frac{1/2}{x + 6}

    Can you finish this?

    -Dan
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  4. #4
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    heres what i've done:

     \int \frac{x + 3}{x^2 + 6 x}dx =

     \int \frac{x}{x^2 + 6 x}dx + \int \frac{3}{x^2 + 6 x}dx

     u = x^2+6x, du = 2x+6

    ln(x+6)+ \int \frac{3}{x^2 + 6 x}dx

    is this right sofar?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Or you can use partial fractions:
    x^2 + 6x = x(x + 6)

    So let's find an A and B such that
    \frac{x + 3}{x^2 + 6x} = \frac{A}{x} + \frac{B}{x + 6}

    Well...
    \frac{A}{x} + \frac{B}{x + 6} = \frac{A(x + 6) + Bx}{x^2 + 6x} = \frac{(A + B)x + 6A}{x^2 + 6x}

    Comparing the two expressions gives me
    A + B = 1
    6A = 3

    Giving me that A = B = \frac{1}{2}.

    So
    \int dx \frac{x + 3}{x^2 + 6x} = \int dx \frac{1/2}{x} + \int dx \frac{1/2}{x + 6}

    Can you finish this?

    -Dan
    That's nice! never thought of doing partial fractions. substitution was the way i immediately saw to do it. and since it's one of the easiest integration techniques, i didn't bother looking for another

    \int \frac {x + 3}{x^2 + 6x}dx

    Let u = x^2 + 6x

    \Rightarrow du = 2x + 6dx = 2(x + 3) dx

    \Rightarrow \frac {1}{2}du = (x + 3)dx

    So our integral becomes:

    \frac {1}{2} \int \frac {1}{u}du
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    heres what i've done:

     \int \frac{x + 3}{x^2 + 6 x}dx =

     \int \frac{x}{x^2 + 6 x}dx + \int \frac{3}{x^2 + 6 x}dx

     u = x^2+6x, du = 2x+6

    ln(x+6)+ \int \frac{3}{x^2 + 6 x}dx

    is this right sofar?
    dont think so. how did you get \ln ( x + 6)?

    you didnt split the integral incorrectly, but i think you're getting yourself into more trouble than it's worth. use topsquark's or my method. do you understand them?
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  7. #7
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    yeah i understand them, needed more practice on it. thanks for the help topsquark & Jhevon
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  8. #8
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    \int {\frac{{x + 3}}<br />
{{x^2 + 6x}}~dx} = \frac{1}<br />
{2}\int {\frac{{\left( {x^2 + 6x} \right)^\prime }}<br />
{{x^2 + 6x}}~dx} = \frac{1}<br />
{2}\ln (x^2 + 6x) + k
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