人邮高数 第3章 第3-2-4题

[AI解答]

[AI解答]

以下是习题3-2第4题各小题的详细解答，均使用分部积分法。

分部积分公式：
$$
\int u \, dv = uv - \int v \, du
$$

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(1) $\displaystyle{\int} x \ln x \, dx$

令 $u = \ln x$，$dv = x \, dx$，则
$du = \frac{1}{x} dx$，$v = \frac{x^2}{2}$
$$
\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx
= \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx
= \frac{x^2}{2} \ln x - \frac{x^2}{4} + C
$$

难度：★☆☆☆☆

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(2) $\displaystyle{\int} \frac{\ln x}{x^2} \, dx$

令 $u = \ln x$，$dv = x^{-2} dx$，则
$du = \frac{1}{x} dx$，$v = -x^{-1}$
$$
\int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} - \int \left(-\frac{1}{x}\right) \cdot \frac{1}{x} dx
= -\frac{\ln x}{x} + \int \frac{1}{x^2} dx
= -\frac{\ln x}{x} - \frac{1}{x} + C
$$

难度：★☆☆☆☆

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(3) $\displaystyle{\int} x e^{-x} \, dx$

令 $u = x$，$dv = e^{-x} dx$，则
$du = dx$，$v = -e^{-x}$
$$
\int x e^{-x} dx = -x e^{-x} + \int e^{-x} dx = -x e^{-x} - e^{-x} + C
$$

难度：★☆☆☆☆

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(4) $\displaystyle{\int} x^2 e^{2x} \, dx$

令 $u = x^2$，$dv = e^{2x} dx$，则
$du = 2x dx$，$v = \frac{1}{2} e^{2x}$
$$
\int x^2 e^{2x} dx = \frac{x^2}{2} e^{2x} - \int x e^{2x} dx
$$
再对 $\int x e^{2x} dx$ 分部：令 $u = x$，$dv = e^{2x} dx$，则
$du = dx$，$v = \frac{1}{2} e^{2x}$
$$
\int x e^{2x} dx = \frac{x}{2} e^{2x} - \frac{1}{2} \int e^{2x} dx = \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x}
$$
代回：
$$
\int x^2 e^{2x} dx = \frac{x^2}{2} e^{2x} - \frac{x}{2} e^{2x} + \frac{1}{4} e^{2x} + C
$$

难度：★★☆☆☆

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(5) $\displaystyle{\int} x \sin x \, dx$

令 $u = x$，$dv = \sin x \, dx$，则
$du = dx$，$v = -\cos x$
$$
\int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C
$$

难度：★☆☆☆☆

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(6) $\displaystyle{\int} x \sin 2x \, dx$

令 $u = x$，$dv = \sin 2x \, dx$，则
$du = dx$，$v = -\frac{1}{2} \cos 2x$
$$
\int x \sin 2x \, dx = -\frac{x}{2} \cos 2x + \frac{1}{2} \int \cos 2x \, dx
= -\frac{x}{2} \cos 2x + \frac{1}{4} \sin 2x + C
$$

难度：★☆☆☆☆

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(7) $\displaystyle{\int} x^2 \cos x \, dx$

令 $u = x^2$，$dv = \cos x \, dx$，则
$du = 2x dx$，$v = \sin x$
$$
\int x^2 \cos x \, dx = x^2 \sin x - 2 \int x \sin x \, dx
$$
由(5)知 $\int x \sin x \, dx = -x \cos x + \sin x$，所以
$$
= x^2 \sin x - 2(-x \cos x + \sin x) + C = x^2 \sin x + 2x \cos x - 2 \sin x + C
$$

难度：★★☆☆☆

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(8) $\displaystyle{\int} x \cos 3x \, dx$

令 $u = x$，$dv = \cos 3x \, dx$，则
$du = dx$，$v = \frac{1}{3} \sin 3x$
$$
\int x \cos 3x \, dx = \frac{x}{3} \sin 3x - \frac{1}{3} \int \sin 3x \, dx
= \frac{x}{3} \sin 3x + \frac{1}{9} \cos 3x + C
$$

难度：★☆☆☆☆

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(9) $\displaystyle{\int} \frac{x}{\cos^2 x} \, dx$

注意 $\frac{1}{\cos^2 x} = \sec^2 x$，令 $u = x$，$dv = \sec^2 x \, dx$，则
$du = dx$，$v = \tan x$
$$
\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx = x \tan x + \ln|\cos x| + C
$$

难度：★★☆☆☆

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(10) $\displaystyle{\int} (x-1) 5^x \, dx$

令 $u = x-1$，$dv = 5^x dx$，则
$du = dx$，$v = \frac{5^x}{\ln 5}$
$$
\int (x-1)5^x dx = \frac{(x-1)5^x}{\ln 5} - \int \frac{5^x}{\ln 5} dx
= \frac{(x-1)5^x}{\ln 5} - \frac{5^x}{(\ln 5)^2} + C
$$

难度：★★☆☆☆

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(11) $\displaystyle{\int} \arcsin x \, dx$

令 $u = \arcsin x$，$dv = dx$，则
$du = \frac{1}{\sqrt{1-x^2}} dx$，$v = x$
$$
\int \arcsin x \, dx = x \arcsin x - \int \frac{x}{\sqrt{1-x^2}} dx
$$
令 $t = 1-x^2$，$dt = -2x dx$，则
$$
\int \frac{x}{\sqrt{1-x^2}} dx = -\sqrt{1-x^2}
$$
所以
$$
\int \arcsin x \, dx = x \arcsin x + \sqrt{1-x^2} + C
$$

难度：★★☆☆☆

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(12) $\displaystyle{\int} \arctan \sqrt{x} \, dx$

令 $t = \sqrt{x}$，则 $x = t^2$，$dx = 2t dt$
$$
\int \arctan \sqrt{x} \, dx = \int \arctan t \cdot 2t \, dt
$$
对 $\int t \arctan t \, dt$ 分部：令 $u = \arctan t$，$dv = t dt$，则
$du = \frac{1}{1+t^2} dt$，$v = \frac{t^2}{2}$
$$
\int t \arctan t \, dt = \frac{t^2}{2} \arctan t - \frac{1}{2} \int \frac{t^2}{1+t^2} dt
= \frac{t^2}{2} \arctan t - \frac{1}{2} \int \left(1 - \frac{1}{1+t^2}\right) dt
$$
$$
= \frac{t^2}{2} \arctan t - \frac{1