# 數學~求積分(分部積分, 有理函數)

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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

• 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