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Calculus

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∫1/(x2 − 2x + 8)3/2 dx

  • Calculus - ,

    Find the following integral ? How , is it by substitution !!!!!!!!

  • Calculus - not sure how but rewrote - ,

    I do not know but suspect you mean

    ∫1/(x^2 − 2x + 8)^(3/2) dx

  • Calculus - ,

    ∫(x2 − 2x + 8)^-1.5 dx

  • Calculus - ,

    http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=7c9206fea2b96c1f22626b03b18327b3&title=Math+Help+Boards:+Indefinite+Integral+Calculator&theme=blue

    says:
    (x-1)/[7 sqrt(x^2-2x+8) ]

  • Calculus - ,

    Note that you have

    ∫1/(x^2 − 2x + 8)^(3/2) dx
    = ∫1/((x-1)^2 + 7)^(3/2) dx
    If x-1 = √7 tanu
    (x-1)^2 + 7 = 7 sec^2 u
    dx = √7 sec^2 u du
    and you have
    ∫1/(√7 secu)^3 * √7 sec^2 u du
    = ∫ 1/7 cosu du
    = 1/7 sinu
    = 1/(7 csc u)
    = 1/(7 √(1+cot^2 u))
    = 1/(7 √(1 + (7/(x-1)^2))
    = 1/(7/(x-1) √((x-1)^2 + 7)
    = 1/ 7√(x^2-2x+8)

    Simpler than I'd have expected.

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