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Math Help - Integration

  1. #1
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    Integration

    Hi there,

    just wondering if someone could help with this: integrate x^3(1+x^8)^-1(dx)

    if u=(x^4)^2

    how does the x^3 become 1 in this equation?
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  2. #2
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    Re: Integration

    \int \frac{x^3}{1+(x^4)^2} \, dx

    u = x^4

    du = 4x^3 \, dx

    \frac{1}{4} \int \frac{4x^3}{1+(x^4)^2} \, dx

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

    can you finish?
    Thanks from NettieL
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  3. #3
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    Re: Integration

    Thank you skeeter! Yes, I understand not that x^3 gets replaced by 4x^3dx (please correct me if that is not the correct explanation), and does (u^2)^-1 become just u (later on) because tan^-1 multiplies and flips it somehow? Not sure I have that right...
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