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Math Help - Integrate with Improper Fractions

  1. #1
    Senior Member polymerase's Avatar
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    Integrate with Improper Fractions

    How do you integrate:

    <br />
\int \dfrac {1}{x^3-8}\,dx<br />

    I got it down to this but i totally stuck on how to integrate one of the improper fraction...

    <br />
\dfrac{-1}{12}\int \dfrac {x+4}{x^2+2x+4}\,dx + \dfrac{1}{2}ln|x-2|<br />

    What do I do.....??
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    How do you integrate:

    <br />
\int \dfrac {1}{x^3-8}\,dx<br />

    I got it down to this but i totally stuck on how to integrate one of the improper fraction...

    <br />
\dfrac{-1}{12}\int \dfrac {x+4}{x^2+2x+4}\,dx + \dfrac{1}{2}ln|x-2|<br />

    What do I do.....??
    \int \frac {x + 4}{x^2 + 2x + 4}~dx = \int \frac {x + 1 + 3}{x^2 + 2x + 4}~dx

    = \int \frac {x + 1}{x^2 + 2x + 4}~dx + \int \frac {3}{x^2 + 2x + 4}~dx

    Now, for the first integral, use the substitution u = x^2 + 2x + 4

    for the second integral, complete the square of the denominator and simplify so you can get the arctan antiderivative (recall \int \frac {1}{x^2 + 1}~dx = \arctan x + C, there is a formula for \int \frac {1}{u^2 + a^2}~dx, but I think practicing the algebraic manipulation would be good for you. then look at the formula)
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  3. #3
    MHF Contributor red_dog's Avatar
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    \displaystyle -\frac{1}{12}\int\frac{x+4}{x^2+2x+4}dx=-\frac{1}{24}\int\frac{2x+2+6}{x^2+2x+4}dx=
    \displaystyle=-\frac{1}{24}\int\frac{2x+2}{x^2+2x+4}dx-\frac{1}{4}\int\frac{1}{(x+1)^2+3}dx=
    \displaystyle=-\frac{1}{24}\ln(x^2+2x+4)-\frac{1}{4\sqrt{3}}\arctan\frac{x+1}{\sqrt{3}}+C
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