习题4.3 分部积分法
1. 填空题
【题目】
$\int x\sin xdx =$ ____；
$\int \ln x dx =$ ____;
$\int \arcsin x dx =$ ____；
$\int e^{\sqrt[3]{x}}dx =$ ____；
5）设 $F'(x) = f(x)$ ，则 $\int xf'(x)dx =$ ____.
【解析】
分部积分法：
$$\int x\sin x dx = -\int x d(\cos x) = -x\cos x + \int \cos x dx = -x\cos x + \sin x + C$$分部积分法：
$$\int \ln x dx = x\ln x - \int x d(\ln x) = x\ln x - \int x \cdot \frac{1}{x}dx = x\ln x - x + C = x(\ln x - 1) + C$$分部积分法：
$$\int \arcsin x dx = x\arcsin x - \int x d(\arcsin x) = x\arcsin x - \int \frac{x}{\sqrt{1 - x^2}}dx$$$$= x\arcsin x + \frac{1}{2}\int (1 - x^2)^{-\frac{1}{2}}d(1 - x^2) = x\arcsin x + \sqrt{1 - x^2} + C$$换元后分部积分：令 $t = \sqrt[3]{x} \implies x = t^3, dx = 3t^2 dt$。
$$\int e^{\sqrt[3]{x}}dx = 3\int t^2 e^t dt = 3\int t^2 d(e^t) = 3\left[t^2 e^t - 2\int t e^t dt\right] = 3\left[t^2 e^t - 2(t e^t - e^t)\right] + C$$$$= 3e^t(t^2 - 2t + 2) + C = 3e^{\sqrt[3]{x}}(\sqrt[3]{x^2} - 2\sqrt[3]{x} + 2) + C$$分部积分法：
$$\int x f'(x)dx = \int x d(f(x)) = x f(x) - \int f(x)dx$$因为 $F'(x) = f(x) \implies \int f(x)dx = F(x) + C$，故：
$$\int x f'(x)dx = x f(x) - F(x) + C$$
2. 计算下列各题
【题目】
$\int xe^{-2x}dx;$
$\int x^{2}\cos xdx;$
$\int \arctan \sqrt{x} dx;$
$\int e^{-x}\cos xdx;$
$\int e^{\sqrt{3x + 9}}dx;$
$\int \cos \ln x dx.$
【解析】
分部积分：
$$\int x e^{-2x}dx = -\frac{1}{2}\int x d(e^{-2x}) = -\frac{1}{2}\left[x e^{-2x} - \int e^{-2x}dx\right] = -\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} + C = -\frac{1}{2}e^{-2x}\left(x + \frac{1}{2}\right) + C$$两次分部积分：
$$\int x^2\cos x dx = \int x^2 d(\sin x) = x^2\sin x - 2\int x\sin x dx$$$$= x^2\sin x - 2[-x\cos x + \sin x] + C = x^2\sin x + 2x\cos x - 2\sin x + C$$分部积分：
$$\int \arctan\sqrt{x} dx = x\arctan\sqrt{x} - \int x d(\arctan\sqrt{x}) = x\arctan\sqrt{x} - \int x \cdot \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}}dx$$$$= x\arctan\sqrt{x} - \frac{1}{2}\int \frac{\sqrt{x}}{1 + x}dx$$令 $t = \sqrt{x}, x = t^2, dx = 2tdt$：
$$\frac{1}{2}\int \frac{t}{1 + t^2} \cdot 2tdt = \int \frac{t^2}{1 + t^2}dt = \int \left(1 - \frac{1}{1 + t^2}\right)dt = t - \arctan t = \sqrt{x} - \arctan\sqrt{x}$$合并得：
$$x\arctan\sqrt{x} - (\sqrt{x} - \arctan\sqrt{x}) + C = (x + 1)\arctan\sqrt{x} - \sqrt{x} + C$$循环分部积分：设 $I = \int e^{-x}\cos x dx$。
$$I = -\int \cos x d(e^{-x}) = -e^{-x}\cos x - \int e^{-x}\sin x dx$$$$= -e^{-x}\cos x + \int \sin x d(e^{-x}) = -e^{-x}\cos x + e^{-x}\sin x - \int e^{-x}\cos x dx = e^{-x}(\sin x - \cos x) - I$$解得：
$$I = \frac{1}{2}e^{-x}(\sin x - \cos x) + C$$换元后分部积分：令 $t = \sqrt{3x + 9} \implies 3x + 9 = t^2 \implies dx = \frac{2}{3}t dt$。
$$\int e^{\sqrt{3x + 9}}dx = \frac{2}{3}\int t e^t dt = \frac{2}{3}(t e^t - e^t) + C = \frac{2}{3}(t - 1)e^t + C = \frac{2}{3}(\sqrt{3x + 9} - 1)e^{\sqrt{3x + 9}} + C$$循环分部积分：令 $t = \ln x \implies x = e^t, dx = e^t dt$。
$$\int \cos\ln x dx = \int e^t \cos t dt$$由指数与余弦乘积的积分标准结论：
$$\int e^t\cos t dt = \frac{e^t}{2}(\cos t + \sin t) + C = \frac{x}{2}(\cos\ln x + \sin\ln x) + C$$