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    #1
    Evaluate the integral.
    $$\displaystyle \int_2^4
    \dfrac{x+2}{x^2+3x-4}\, dx$$
    W|A returned these partial fractions but I don't know where the the A=2 B=3 and 5 came from
    $$
    \dfrac{x+2}{(x+4)(x-1)}
    =\dfrac{2}{5(x+4)}+\dfrac{3}{5(x-1)}$$

    the book answer was
    $$\dfrac{4}{5}\ln{2}+\dfrac{1}{5}\ln {3}
    =\dfrac{1}{5}\ln{48}$$
    Last edited by karush; October 8th, 2019 at 15:50.

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    #2
    $\dfrac{x+2}{x^2+3x-4} = \dfrac{x+2}{(x+4)(x-1)} = \dfrac{A}{x+4} + \dfrac{B}{x-1}$

    $ \dfrac{x+2}{(x+4)(x-1)} = \dfrac{A(x-1)}{(x+4)(x-1)} + \dfrac{B(x+4)}{(x+4)(x-1)}$

    common denominator for all terms $\implies$ numerators form the equation ...


    $x+2 = A(x-1) + B(x+4)$

    two methods ...

    (1) matching coefficients

    $x+2 = (A+B)x + (4B-A) \implies A+B = 1 \text{ and } 4B-A = 2$

    solving the system by elimination yields $5B = 3 \implies B = \dfrac{3}{5} \text{ and } A = \dfrac{2}{5}$

    $ \dfrac{x+2}{(x+4)(x-1)} = \dfrac{2}{5(x+4)} + \dfrac{3}{5(x-1)}$


    (2) Heaviside method ... pick strategic values for $x$ to eliminate $A$ or $B$

    $x+2 = A(x-1) + B(x+4)$

    let $x=1$ ...

    $1+2 = A(1-1) + B(1+4) \implies B = \dfrac{3}{5}$

    let $x=-4$ ...

    $-4+2 = A(-4-1) + B(-4+4) \implies A = \dfrac{2}{5}$


    $ \displaystyle \int \dfrac{2}{5(x+4)} + \dfrac{3}{5(x-1)} \, dx = \dfrac{2}{5}\ln(x+4) + \dfrac{3}{5}\ln(x-1) = \dfrac{\ln[(x+4)^2(x-1)^3]}{5}$


    $\bigg[ \dfrac{\ln[(x+4)^2(x-1)^3]}{5} \bigg]_2^4 = \dfrac{\ln(64 \cdot 27) - \ln(36)}{5} = \dfrac{\ln(48)}{5}$

  4. MHB Master
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    #3 Thread Author
    OK I see

    you let on of the variables be multiplied by zero to isolate the the other one


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    #4 Thread Author
    wow with 377 views that was a great help

    they did one in class but I got lost on getting A and B

    really appreciate the steps

    if this were yahoo it would be torture to read without the latex

    Mahalo

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    #5
    Quote Originally Posted by karush View Post
    wow with 377 views that was a great help

    they did one in class but I got lost on getting A and B

    really appreciate the steps

    if this were yahoo it would be torture to read without the latex

    Mahalo
    Yeah but I still sometimes miss those old days where DOS was high tech and Windows hadn't been invented yet.

    -Dan

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