第四章 自测题
计算下列不定积分
【题目】
$\int \frac{1}{\sqrt[3]{2 - 3x}} dx$
$\int \frac{1}{x\ln x\ln\ln x} dx$
$\int \sin 2x\cos 3xdx$
$\int \frac{\arctan\sqrt{x}}{\sqrt{x}(1 + x)} dx$
$\int \frac{1 + \ln x}{(x\ln x)^2} dx$
$\int \frac{1}{\sqrt{x}}\tan \sqrt{x}\sec^3\sqrt{x} dx$
$\int \frac{\sqrt[3]{x}}{x\left(\sqrt{x} + \sqrt[3]{x}\right)} dx$
$\int \sin \sqrt{x} dx;$
$\int \frac{1}{x + \sqrt{1 - x^2}} dx.$
【解析】
凑微分：
$$\int (2 - 3x)^{-\frac{1}{3}}dx = -\frac{1}{3}\int (2 - 3x)^{-\frac{1}{3}}d(2 - 3x) = -\frac{1}{3} \cdot \frac{3}{2}(2 - 3x)^{\frac{2}{3}} + C = -\frac{1}{2}(2 - 3x)^{\frac{2}{3}} + C$$逐层凑微分：
$$\int \frac{1}{x\ln x\ln\ln x}dx = \int \frac{d(\ln x)}{\ln x \ln(\ln x)} = \int \frac{d(\ln\ln x)}{\ln\ln x} = \ln|\ln\ln x| + C$$积化和差公式 $\sin 2x\cos 3x = \frac{1}{2}[\sin 5x - \sin x]$：
$$\int \sin 2x\cos 3x dx = \frac{1}{2}\int (\sin 5x - \sin x)dx = -\frac{1}{10}\cos 5x + \frac{1}{2}\cos x + C = \frac{1}{2}\cos x - \frac{1}{10}\cos 5x + C$$凑微分：$d(\arctan\sqrt{x}) = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}}dx = \frac{dx}{2\sqrt{x}(1 + x)}$。
$$\int \frac{\arctan\sqrt{x}}{\sqrt{x}(1 + x)}dx = 2\int \arctan\sqrt{x} d(\arctan\sqrt{x}) = (\arctan\sqrt{x})^2 + C$$观察微分：$d(x\ln x) = (\ln x + 1)dx$。
$$\int \frac{1 + \ln x}{(x\ln x)^2}dx = \int (x\ln x)^{-2}d(x\ln x) = -(x\ln x)^{-1} + C = -\frac{1}{x\ln x} + C$$凑微分：令 $t = \sqrt{x}, dt = \frac{dx}{2\sqrt{x}}$。
$$\int \frac{1}{\sqrt{x}}\tan\sqrt{x}\sec^3\sqrt{x}dx = 2\int \sec^2 t \cdot \sec t\tan t dt = 2\int \sec^2 t d(\sec t) = \frac{2}{3}\sec^3 t + C = \frac{2}{3}\sec^3\sqrt{x} + C$$化简被积函数并换元：分子分母同除以 $\sqrt[3]{x}$ 得 $\frac{1}{x(x^{\frac{1}{6}} + 1)}$。 令 $t = \sqrt[6]{x} \implies x = t^6, dx = 6t^5 dt$：
$$\int \frac{6t^5}{t^6(t + 1)}dt = 6\int \frac{1}{t(t + 1)}dt = 6\int \left(\frac{1}{t} - \frac{1}{t + 1}\right)dt = 6(\ln|t| - \ln|t + 1|) + C$$因为 $6\ln|t| = \ln|t^6| = \ln x$，所以：
$$\ln x - 6\ln(1 + \sqrt[6]{x}) + C$$换元后分部积分：令 $t = \sqrt{x} \implies x = t^2, dx = 2tdt$。
$$\int \sin\sqrt{x} dx = 2\int t\sin t dt = -2\int t d(\cos t) = -2[t\cos t - \sin t] + C = -2\sqrt{x}\cos\sqrt{x} + 2\sin\sqrt{x} + C$$三角代换：令 $x = \sin t\ (t \in (-\frac{\pi}{4}, \frac{\pi}{2}))$，则 $dx = \cos t dt$，$\sqrt{1 - x^2} = \cos t$。
$$\int \frac{\cos t}{\sin t + \cos t}dt$$利用对称拆项法：
$$\cos t = \frac{1}{2}[(\cos t + \sin t) + (\cos t - \sin t)]$$$$\int \frac{\cos t}{\sin t + \cos t}dt = \frac{1}{2}\int 1 dt + \frac{1}{2}\int \frac{\cos t - \sin t}{\sin t + \cos t}dt = \frac{1}{2}t + \frac{1}{2}\ln|\sin t + \cos t| + C$$代回 $t = \arcsin x$：
$$\frac{1}{2}\left(\arcsin x + \ln|x + \sqrt{1 - x^2}|\right) + C$$