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Math Help - Evaluate the integral?

  1. #1
    Super Member fardeen_gen's Avatar
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    Evaluate the integral?

    Evaluate:
    \int \frac{dx}{(x^6 - a^6 + a^2x^4 - a^4x^2)}
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  2. #2
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    It might help to know:

    x^6-a^6+a^2*x^4-a^4*x^2=(x^4-a^4)(x^2+a^2)
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  3. #3
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by sirsosay@gmail.com View Post
    It might help to know:

    x^6-a^6+a^2*x^4-a^4*x^2=(x^4-a^4)(x^2+a^2)
    Thank you!
    I am still unable to get this:
    Answer:
    Spoiler:
    \frac{1}{8a^5}\ln\left|\frac{x - a}{x + a}\right| - \frac{1}{4a^4}\frac{x}{(x^2 + a^2)} - \frac{1}{2a^5}\arctan\left(\frac{x}{a}\right) + C
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  4. #4
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    \begin{aligned}<br />
(x^4-a^4)(x^2+a^2)&=(x^2-a^2)(x^2+a^2)(x^2+a^2) \\<br />
&=(x-a)(x+a)(x^2+a^2)^2 \end{aligned}

    Then partial fraction decomposition :

    find b,c,d,e such that :
    \frac{1}{(x-a)(x+a)(x^2+a^2)^2}=\frac{b}{x-a}+\frac{c}{x+a}+\frac{d}{x^2+a^2}+\frac{e}{(x^2+a  ^2)^2}

    You should get b=\frac{1}{8a^5} ~,~ c=\frac{-1}{8a^5} ~,~ d=\frac{-1}{4a^4} ~,~ e=\frac{-1}{2a^2}
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