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Math Help - Intergral Help!

  1. #1
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    Exclamation Intergral Help!

    Im struggling to find the integral of the following:

    2
    ____________

    (x+2)^2 (2-x)

    Can someone show me the step by step solution. Plz thanks
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  2. #2
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    Quote Originally Posted by Nimmy
    Im struggling to find the integral of the following:

    2
    ____________

    (x+2)^2 (2-x)

    Can someone show me the step by step solution. Plz thanks
    I presume you mean
    \int \frac{2dx}{(x+2)^2(2-x)}
    This type of problem requires fractional decomposition.
    Thus, express
    \frac{2}{(x+2)^2(2-x)}=\frac{A}{(x+2)}+\frac{B}{(x+2)^2}+\frac{C}{(2-x)}
    Now multiply the identity by (x+2)^2(2-x)
    Thus,
    2=A(x+2)(2-x)+B(2-x)+C(x+2)^2
    Open sesame,
    2=x^2(-A+C)+x(4C-B)+(4A+2B+4C)
    Now that happens when
    -A+C=0
    4C-B=0
    4A+2B+4C=2
    Solving this we get that,
    (A,B,C)=(1/8,1/2,1/8)
    Thus,
    \frac{2}{(x+2)^2(2-x)}=\frac{1/8}{x+2}+\frac{1/2}{(x+2)^2}+\frac{1/8}{2-x}
    Thus,
    \int \frac{2dx}{(x+2)^2(2-x)}=\int (\frac{1/8}{x+2}+\frac{1/2}{(x+2)^2}+\frac{1/8}{2-x}) dx
    But,
    \int (\frac{1/8}{x+2}+\frac{1/2}{(x+2)^2}+\frac{1/8}{2-x}) dx= \frac{1}{8}\ln |x+2|+\frac{1}{8}\ln |x-2|-\frac{1}{2}(x+2)^{-1}+C
    Q.E.D.

    Note it is possible I made a mistake somewhere I am so tired right now and it took me a long time to write this in LaTeX form.
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