# 高等数学——砍瓜切菜算积分的分部积分法

2019/04/10 10:10

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## 原理和推导

$$(uv)'=u'v + uv'$$

$$uv' = (uv)' - u'v$$

$$\int uv'dx = uv - \int u'vdx$$

$$\int udv = uv - \int vdu$$

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## u和v的选择

$$\int x \cos x dx = x \sin x - \int \sin x dx$$

$\int \sin x dx$很容易得到原函数，所以整体的答案就是：

$$\int x \cos x dx = x \sin x + \cos x + C$$

$$\int x \cos x dx = \cos x \cdot \frac{x^2}{2} + \int \frac{x^2}{2}\sin x dx$$

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## 一点诀窍

$$\int x e^x dx$$

$$\int x e^x dx = xe^x - \int e^x dx = x e^x - e^x + C$$

$$\int x \ln x dx$$

\begin{aligned} \int x \ln x dx &= \frac{x^2}{2} \ln x - \int \frac{1}{x}\cdot \frac{x^2}{2}dx \ &=\frac{x^2}{2} - \int \frac{x}{2} dx \ &=\frac{x^2}{2} - \frac{x^2}{4} + C \end{aligned}

$$\int x^2 e^x dx$$

$$\int x^2 e^x dx = x^2e^x - \int 2x e^x dx$$

\begin{aligned} \int x^2 e^x dx &= x^2e^x - \int 2x e^x dx\ &=x^2 e^x - 2x \cdot e^x + \int 2 e^x dx \ &= e^x (x^2 - 2x + 2) \end{aligned}

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## 和换元法结合

$$\int e^{\sqrt{x}}dx$$

$$\int e^{\sqrt{x}}dx= 2\int t e^t dt$$

$$2\int t e^t dt = 2(t e^t - \int e^t dt) = 2e^t(t - 1) = 2e^{\sqrt{x}}(\sqrt{x} - 1) + C$$

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