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Math Help - partial fractions

  1. #1
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    partial fractions

    find the antiderivative of:

    (x^3 - 5x^2 - 4) / (x^4 - 4x^2)

    i used the difference of two squares to get it into this form

    (x^3 - 5x^2 - 4) = (Ax + B)(x^2 - 2x) + (Cx + D)(x^2 - 2x)

    but i can't seem to solve for A,B,C,D and I'm not getting any term equal to the -4 on the left side, what am i doing wrong

    can someone post a step by step solution from this point onwards?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by razorfever View Post
    find the antiderivative of:

    (x^3 - 5x^2 - 4) / (x^4 - 4x^2)

    i used the difference of two squares to get it into this form

    (x^3 - 5x^2 - 4) = (Ax + B)(x^2 - 2x) + (Cx + D)(x^2 - 2x)

    but i can't seem to solve for A,B,C,D and I'm not getting any term equal to the -4 on the left side, what am i doing wrong

    can someone post a step by step solution from this point onwards?

    try this factorization for the denominator

    (x^4-4x^2)=x^2(x^2-4)=x^2(x-2)(x+2)
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  3. #3
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    oh nice
    but why doesn't the one i'm doing work?
    what's wrong with it?
    it seems like the more obvious way to go right?
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  4. #4
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    \int{\frac{x^{3}-5x^{2}-4}{x^{4}-4x^{2}}\,dx}=\underbrace{\int{\frac{x}{x^{2}-4}\,dx}}_{\alpha }\,-\,5\underbrace{\int{\frac{dx}{x^{2}-4}}}_{\beta }\,-\,4\underbrace{\int{\frac{dx}{x^{4}-4x^{2}}}}_{\gamma }.

    \alpha is easy to solve, so I'll take care of \gamma. Put x=\frac1u and the integral is

    \begin{aligned}<br />
   \int{\frac{dx}{x^{4}-4x^{2}}}&=\int{\frac{u^{2}}{4u^{2}-1}\,du} \\ <br />
 & =\frac{1}{4}\int{\frac{4u^{2}-1+1}{4u^{2}-1}\,du} \\ <br />
 & =\frac{1}{4}\left\{ \int{du}+\underbrace{\int{\frac{du}{(2u+1)(2u-1)}}}_{\varphi } \right\}.<br />
\end{aligned}

    Now I'll solve \beta, thus we have \int{\frac{dx}{x^{2}-4}}=\frac{1}{8}\int{\frac{(x+4)-(x-4)}{(x+4)(x-4)}\,dx}=\frac{1}{8}\left( \int{\frac{dx}{x-4}}-\int{\frac{dx}{x+4}} \right). Finally, by using this procedure, you can easily solve \varphi, thus, no more integrals and we're done.
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