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Math Help - Integral 01

  1. #1
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    Integral 01

    \int \frac{dx}{x^4\sqrt{1+x^2}}



    Answer:
    \frac{(2x^2-1)(1+x^2)\frac{1}{2}}{3x^3}
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  2. #2
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    \int \frac{dx}{x^4\sqrt{1+x^2}}

    You can use trigonometric subtitution.

    Let:
    x = \tan{\theta}

    dx = \sec^2{\theta} d\theta


    \int \frac{\sec^2{\theta}}{(\tan{\theta})^4\sqrt{1+\tan  ^2{\theta}}} d\theta

    \int \frac{\sec^2{\theta}}{(\tan{\theta})^4\sec{\theta}  } d\theta

    \int \frac{\sec{\theta}}{(\tan{\theta})^4} d\theta

    \int \frac{1}{\cos{\theta}} \cdot \frac{cos^4{\theta}}{\sin^4{\theta}} d\theta

    \int \frac{cos^3{\theta}}{\sin^4{\theta}} d\theta

    \int \frac{(1-sin^2{\theta})\cos{\theta}}{\sin^4{\theta}} d\theta

    Let
    u = \sin{\theta}

    du = \cos{\theta} d\theta

    \int \frac{1-u^2}{u^4} du

    Can you continue?
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  3. #3
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    Krizalid's Avatar
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    Think in a reciprocal substitution, like u=\frac1x, the integral equals -\int{\frac{u^{3}}{\sqrt{1+u^{2}}}\,du}. Now put z^2=1+u^2.
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  4. #4
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    Look my solution


    \int \frac{dx}{x^4\sqrt{1+x^2}} = \int x^(-4) (1+x^2)^-\frac{1}{2}dx

    1+x^2=z^2
    x^2=z^2-1
    dx= \frac{zdz}{\sqrt{z^2-1}}


    \int (z^2+1)^(-2) (z^2)^\frac{1}{2}\frac{z}{\sqrt{z^2-1}}
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