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阿阿 黃 發問時間: 科學數學 · 6 年前

∫(1→9) 1 / (1+√t )dt

原題目:

f(x) = ∫(x→x^3+8) x / (1+√t) dt .Find f(1) = ?

1 個解答

評分
  • 6 年前
    最佳解答

    令 t = tan^4 θ,所以 dt = 4 tan^3 θ sec^2 θ dθ

    當 t = 1 時,θ = π/4;當 t = 9 時,θ = π/3;所以

    f(1)

    = ∫(1→9) 1 / (1 + √t) dt

    = ∫(π/4→π/3) 4 tan^3 θ sec^2 θ / (1 + tan^2 θ) dθ

    = ∫(π/4→π/3) 4 tan^3 θ dθ

    = ∫(π/4→π/3) 4 tan θ sec^2 θ dθ - ∫(π/4→π/3) 4 tan θ dθ

    = ∫(π/4→π/3) 4 tan θ d(tan θ) - 4 ∫(π/4→π/3) tan θ dθ

    = (π/4→π/3) [2 tan^2 θ - 4 ln(tan θ + sec θ)]

    = [6 - 4 ln(2 + √3)] - [2 - 4 ln(1 + √2)]

    = 4 + 4 ln[(1 + √2)/(2 + √3)]

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