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匿名使用者 發問時間: 科學數學 · 1 0 年前

數學~求積分(分部積分, 有理函數)

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分部積分, 有理函數

請教下怎樣做

已更新項目:

答案有一些不同

1(2) (1/8)x^4 (2In^2 x - Inx + 1/4)+C

2(2) -(3e^-2 + e^-2)

2(3) In2 - 1/2

1 個解答

評分
  • 1 0 年前
    最佳解答

    <A>

    1.

    ∫ ln(x)x^(-2) dx =(-1)∫ ln(x) d(x^(-1))

    _____________= (-1)[ln(x)x^(-1) - ∫x^(-1)d(ln(x))]

    _____________= (-1)[ln(x)x^(-1) -∫x^(-2)dx]

    _____________= (-1)[ln(x)x^(-1) +x^(-1)]

    2.

    ∫x^3{[ln(x)]^2}dx = (1/4)∫[ln(x)]^2 d(x^4)

    _______________=(1/4){(x^4){[ln(x)]^2}- ∫(x^4) d(ln(x))^2}

    _______________=(1/4){(x^4){[ln(x)]^2}- ∫2(x^3)ln(x) dx}

    _______________=(1/4){(x^4){[ln(x)]^2}- ∫2(x^3)ln(x) dx}

    _______________=(1/4){(x^4){[ln(x)]^2}- (1/2)∫ln(x) d(x^4)}

    _______________=(1/4){(x^4){[ln(x)]^2}- (1/2)ln(x)(x^4)+(1/2)∫(x^4)d(ln(x))}

    _______________=(1/4){(x^4){[ln(x)]^2}- (1/2)ln(x)(x^4)+(1/2)∫(x^3)dx}

    _______________=(1/4){(x^4){[ln(x)]^2}- (1/2)ln(x)(x^4)+(1/8)(x^4)}

    4.

    ∫(x^5) exp(x) dx = (1/D)(x^5) exp(x) p.s.<D=(d/dx)>

    ______________= exp(x)[1/(D+1)](x^5)

    ______________= exp(x)Σ[n=0~∞]((-1)^n)(D^n)(x^5)

    ______________= exp(x)Σ[n=0~∞]((-1)^n)(D^n)(x^5)

    ______________= exp(x){(x^5)-5(x^4)+20(x^3)-60(x^2)+120(x^1)-120}

    <B>

    1.

    ∫[x=1~exp] ln(x) dx = {x*ln(x) - ∫x d[ln(x)]}|[x=1~exp]

    ________________= {exp - ∫1 dx}|[x=1~exp]

    ________________= exp -(exp -1)

    ________________= 1

    2.

    ∫[x=-2~2] x*exp(-x) dx = -∫[x=-2~2] xd[exp(-x)]

    ___________________= -∫[x=-2~2] xd[exp(-x)]

    ___________________= {-x*exp(-x) +∫exp(-x)dx]}|[x=-2~2]

    ___________________= {-x*exp(-x)-exp(-x)}|[x=-2~2]

    ___________________= - (x+1)*exp(-x)|[x=-2~2]

    3.

    _∫[x=0~√ln(2)](x^3)exp(x^2) dx

    =(1/2)∫[x=0~√ln(2)](x^2)exp(x^2) d(x^2)

    =(1/2)∫[x=0~√ln(2)](x^2) d[exp(x^2)]

    =(1/2){(x^2)exp(x^2) -∫exp(x^2) d(x^2)} | [x=0~√ln(2)]

    =(1/2){(x^2)exp(x^2) - exp(x^2)} | [x=0~√ln(2)]

    =(1/2)exp(x^2){(x^2) - 1} | [x=0~√ln(2)]

    <C>

    1.

    ∫1/[(x^2)-x-6] dx = ∫1/[(x^2)-x-6] dx

    ______________= ∫1/[(x+2)(x-3)] dx

    ______________= (1/5)∫[-1/(x+2)]+[1/(x-3)] dx

    ______________=(1/5)∫[-1/(x+2)]dx+∫[1/(x-3)] dx

    ______________=(1/5){∫[-1/(x+2)]d(x+2)+∫[1/(x-3)] d(x-3)}

    ______________=(1/5){-ln(x+2)+ln(x-3)}

    ______________=(1/5)ln[(x-3)/(x+2)]

    2.

    ∫[(x^4)-6(x^2)-7]/(x^3) dx =∫x-6[x^(-1)]-7[x^(-3)] dx

    ___________________=(1/2)(x^2)-6ln(x)+(7/2)[x^(-2)] dx

    2008-12-04 23:00:12 補充:

    (2) --->我的解答提出(1/2)(x^4),即為你的答案

    <b>(2) --->我的解答帶入[x=-2~2]上下限,即為你的答案

    <b>(3)---->我的解答帶入[x=0~√ln(2)]上下限,即為你的答案</b></b>

    參考資料: me
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