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Math Help - Integral

  1. #1
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    Integral

    (sorry I am not that good with latex)

    How do I solve the integral, dx/(1+x^3+sqrt(1+x^6)), from -1 to 1?
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  2. #2
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    Quote Originally Posted by Jake8054 View Post
    (sorry I am not that good with latex)

    How do I solve the integral, dx/(1+x^3+sqrt(1+x^6)), from -1 to 1?
    If the integral is

    \int_{-1}^1 \frac{dx}{1+x^3 + \sqrt{1+x^6}}

    then first rationalize the denominator

    \frac{1}{2} \int_{-1}^1 \frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx

    Then split the integral

    \frac{1}{2} \left( \int_{-1}^0 \frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)

    Then make the substitution  x = -u in the first integral

    \frac{1}{2} \left(- \int_{1}^0 \frac{ 1 - u^3 - \sqrt{1+u^6} }{- u^3 }\, du + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)

    or

    \frac{1}{2} \left( -\int_{0}^1 \frac{ 1 - u^3 - \sqrt{1+u^6} }{ u^3 }\, du + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)

    since u is a dummy variable then

    \frac{1}{2} \left( -\int_{0}^1 \frac{ 1 - x^3 - \sqrt{1+x^6} }{ x^3 }\, dx + \int_{0}^1\frac{ 1 + x^3 - \sqrt{1+x^6} }{x^3 }\, dx \right)

    so

    \frac{1}{2} \int_{0}^1 \frac{ 2x^3 }{ x^3 }\, dx = 1
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  3. #3
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    zomg i cant believe it was that simple. Thank you so much!
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