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Опубликовано 5 лет назад по предмету Математика от sergejpetrakov

помогите решить интегрирование по частям int sqrtx lnx dx

  1. Ответ
    Ответ дан BuTaC3

    ∫ √x ln x dx =

    √x = t →
    x = t² → dx = 2t dt
    ∫ √x ln x dx = ∫ t ln (t²) (2t dt) =
    ∫ 2t² ln (t²) dt =
    (t²) = u and 2t²dt = dv;
    (2t / t²) dt = (2 / t) dt = du

     2(t²⁺¹)/(2+1) = (2/3)t³ = v


    ∫ 2t² ln (t²) dt = (2/3)t³ ln (t²) - ∫ (2/3)t³ (2/ t) dt =
    (2/3)t³ ln (t²) - (4/3) ∫ (t³/ t) dt =
    (2/3)t³ ln (t²) - (4/3) ∫ t² dt =
    (2/3)t³ ln (t²) - (4/3)(t²⁺¹)/(2+1) + c =
    (2/3)t³ ln (t²) - (4/3)(1/3)t³+ c =
    (2/3)t³ ln (t²) - (4/9)t³+ c 
     t = √x
    ∫ √x ln x dx = (2/3)√x³ ln (√x²) - (4/9)√x³+ c = 
    (2/3)x√x ln x - (4/9)x√x + c

  2. Ответ
    Ответ дан PhysM

    int{sqrt{x}lnx}, dx

    U=lnx

    dU=frac{1}{x}dx

    V=frac{2}{3}*sqrt{x^3}

    int{sqrt{x}lnx}, dx=frac{2}{3}*sqrt{x^3}lnx-int{frac{2}{3}sqrt{x^3}}, dx=frac{2}{3}*sqrt{x^3}lnx-int{frac{2}{3}sqrt{x^3}frac{1}{x}}, dx=frac{2}{3}*sqrt{x^3}lnx-frac{4}{9}sqrt{x^3}+C