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  1. #1
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    integral

    evalute: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$
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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: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$
    use substitution. u = x^2 + 6x
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by viet View Post
    evalute: $\displaystyle \int \frac{x + 3}{x^2 + 6 x}dx$
    Or you can use partial fractions:
    $\displaystyle x^2 + 6x = x(x + 6)$

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

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

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

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

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

    Can you finish this?

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

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

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

    $\displaystyle u = x^2+6x, du = 2x+6$

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

    is this right sofar?
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    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:
    $\displaystyle x^2 + 6x = x(x + 6)$

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

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

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

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

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

    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

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

    Let $\displaystyle u = x^2 + 6x$

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

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

    So our integral becomes:

    $\displaystyle \frac {1}{2} \int \frac {1}{u}du$
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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
    heres what i've done:

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

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

    $\displaystyle u = x^2+6x, du = 2x+6$

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

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

    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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    $\displaystyle \int {\frac{{x + 3}}
    {{x^2 + 6x}}~dx} = \frac{1}
    {2}\int {\frac{{\left( {x^2 + 6x} \right)^\prime }}
    {{x^2 + 6x}}~dx} = \frac{1}
    {2}\ln (x^2 + 6x) + k$
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