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Math Help - partial fractions - rationalizing substitutions

  1. #1
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    partial fractions - rationalizing substitutions

    can someone help with the steps underlined, how do you i get from one to the other?

    Thanks.
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  2. #2
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    Quote Originally Posted by calculus0 View Post
    can someone help with the steps underlined, how do you i get from one to the other?

    Thanks.
    For this one..

    \frac{u^2}{u^2-4}=\frac{u^2-4+4}{u^2-4}=\frac{u^2-4}{u^2-4}+\frac{4}{u^2-4}=1+\frac{4}{u^2-4}
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    Quote Originally Posted by calculus0 View Post
    can someone help with the steps underlined, how do you i get from one to the other?

    Thanks.
    2 \int \frac{u^2}{u^2-4} = 2 \int \frac{u^2 -4 +4}{u^2-4} =  2 \int \frac{(u^2 -4) +4}{u^2-4}

    =  2 \int [\frac{u^2 -4}{u^2-4} + \frac{4}{u^2-4}] =  2 \int [1+ \frac{4}{u^2-4}]
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    so are you essentially adding and subtracting 4? (how would you know to do that?
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    Quote Originally Posted by calculus0 View Post
    so are you essentially adding and subtracting 4? (how would you know to do that?
    When you have \frac{u^2}{u^2-k}

    this can be rearranged to a form which can be directly integrated
    using the tables of standard known integrals

    by adding and subtracting k in the numerator, whatever constant k is.

    \frac{u^2}{u^2-k}=\frac{u^2-k+k}{u^2-k}=1+\frac{k}{u^2-k}

    both of which which can easily be integrated.
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    More generally, just do a "long division"

    u^2- 0u- 4)\overline{u^2+ 0u+ 0}

    Obviously the " u^2" will divide into the " u^2" 1 time so we will then need to subtract (u^2+ 0u+ 0)- (u^2- 0u- 4)= 4. u^2- 4 divides into u^2 1 time with remainder 4: \frac{u^2}{u^2- 4}= 1+ \frac{4}{u^2- 4}.
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