第二节 不定积分的计算

基本积分公式

现阶段常用公式如下:

(1) $\int x^{\alpha} \mathrm{d} x=\frac{x^{\alpha+1}}{\alpha+1}+C(\alpha \neq-1$, 为常数 $) \quad$

(2) $\int \frac{1}{x} \mathrm{~d} x=\ln |x|+C$

(3) $\int a^{x} \mathrm{~d} x=\frac{1}{\ln a} a^{x}+C(a>0, a \neq 1) \quad$

(4) $\int \mathrm{e}^{x} \mathrm{~d} x=\mathrm{e}^{x}+C$

(5) $\int \sin x \mathrm{~d} x=-\cos x+C$

(6) $\int \cos x \mathrm{~d} x=\sin x+C$

(7) $\int \sec ^{2} x \mathrm{~d} x=\int \frac{1}{\cos ^{2} x} \mathrm{~d} x=\tan x+C$

(8) $\int \csc ^{2} x \mathrm{~d} x=\int \frac{1}{\sin ^{2} x} \mathrm{~d} x=-\cot x+C$

(9) $\int \frac{\mathrm{d} x}{\sqrt{1-x^{2}}}=\arcsin x+C$

(10) $\int \frac{\mathrm{d} x}{1+x^{2}}= \arctan x+C$

计算下列不定积分

(1)$\int\frac{1}{\sqrt{x}}\mathrm{~d} x$

(2) $\int x^{2}\cdot\sqrt[3]{x}\mathrm{~d} x$

(3) $\displaystyle\int \dfrac{3 x^{4}+2 x^{2}}{x^{2}+1} \mathrm{~d} x$.

▶答案

(1) $2\sqrt{x} + C$

(2) $\frac{3}{10}x^{\frac{10}{3}} + C$

(3) $x^3 - x + \arctan x + C$

▶详解

【解】 (1)

$$\begin{aligned} \text{原式} &= \int x^{-\frac{1}{2}}\mathrm{~d}x \\ &= 2\sqrt{x} + C. \end{aligned}$$

(2)

$$\begin{aligned} \text{原式} &= \int x^{\frac{7}{3}}\mathrm{~d}x \\ &= \frac{3}{10}x^{\frac{10}{3}} + C. \end{aligned}$$

(3)对于分子次数不小于分母次数的有理函数积分, 关键在于使分子多项式次数小于分母 , 最终利用基本积分公式 $\int \frac{1}{1+x^2}\mathrm{~d}x = \arctan x + C$ 求解.

$$\begin{aligned} \text{原式} &= \int \frac{3x^2(x^2+1) - (x^2+1) + 1}{x^2+1} \mathrm{~d}x \\ &= \int \left( 3x^2 - 1 + \frac{1}{x^2+1} \right) \mathrm{~d}x \\ &= x^3 - x + \arctan x + C. \end{aligned}$$

计算不定积分

(1)$\displaystyle\int \dfrac{\mathrm{d} x}{x^{2} \sqrt{x}}$.

(2)$\displaystyle\int\left(\dfrac{3}{1+x^{2}}-\dfrac{2}{\sqrt{1-x^{2}}}\right) \mathrm{d} x$.

(3) $\int \frac{(x-2)^{2}}{x} \mathrm{~d} x$

▶提示

(1)利用指数运算法则将根式转化为幂函数进行计算;

(2)拆分后应用基本积分公式;

(3)完全平方展开后拆分,将分子次数降低.

▶答案

(1) $-\frac{2}{3}x^{-\frac{3}{2}} + C$

(2) $3 \arctan x - 2 \arcsin x + C$

(3) $\frac{1}{2}x^2 - 4x + 4\ln|x| + C$

▶详解

【解】 (1) 将被积函数化为幂函数形式, 直接利用幂函数积分公式可得

$$\begin{aligned} \int \frac{\mathrm{d} x}{x^{2} \sqrt{x}} &= \int x^{-\frac{5}{2}} \mathrm{~d} x \\ &= -\frac{2}{3}x^{-\frac{3}{2}} + C \end{aligned}$$

(2) 利用积分的线性性质及基本积分公式, 直接求解可得

$$\begin{aligned} \int\left(\frac{3}{1+x^{2}}-\frac{2}{\sqrt{1-x^{2}}}\right) \mathrm{d} x &= 3 \int \frac{1}{1+x^{2}} \mathrm{~d} x - 2 \int \frac{1}{\sqrt{1-x^{2} }} \mathrm{~d} x \\ &= 3 \arctan x - 2 \arcsin x + C. \end{aligned}$$

(3) 将被积函数的分子展开, 并逐项除以分母, 得

$$\begin{aligned} \int \frac{(x-2)^{2}}{x} \mathrm{~d} x &= \int \frac{x^2 - 4x + 4}{x} \mathrm{~d} x \\ &= \int \left( x - 4 + \frac{4}{x} \right) \mathrm{~d} x \\ &= \frac{1}{2}x^2 - 4x + 4\ln|x| + C. \end{aligned}$$

计算不定积分 $\int \cot^2 x \mathrm{~d}x$.

▶提示

利用恒等式 $\cot^2 x = \csc^2 x - 1$ 将被积函数化简, 求积分时需特别注意 $\csc^2 x$ 的原函数带负号.

▶答案

$-\cot x - x + C$

▶详解

【解】

$$\begin{aligned} \int \cot^2 x \mathrm{~d}x &= \int (\csc^2 x - 1) \mathrm{~d}x \\ &= -\cot x - x + C. \end{aligned}$$

计算不定积分 $\int \tan^2 x \mathrm{~d}x$.

▶提示

利用恒等式 $\tan^2 x = \sec^2 x - 1$ 变形.

▶答案

$\tan x - x + C$

▶详解

【解】

$$\begin{aligned} \int \tan^2 x \mathrm{~d}x &= \int (\sec^2 x - 1) \mathrm{~d}x \\ &= \tan x - x + C. \end{aligned}$$

计算不定积分 $\int \sin^2 x \mathrm{~d}x$.

▶提示

$\sin x$偶次幂的积分, 使用半角公式(降幂公式)$\sin^2 x = \frac{1 - \cos 2x}{2}$化简计算.

▶答案

$\frac{1}{2}x - \frac{1}{4}\sin 2x + C$

▶详解

【解】

$$\begin{aligned} \int \sin^2 x \mathrm{~d}x &= \int \frac{1 - \cos 2x}{2} \mathrm{~d}x \\ &= \frac{1}{2}x - \frac{1}{4}\sin 2x + C. \end{aligned}$$

计算不定积分 $\int \cos^2 x \mathrm{~d}x$.

▶提示

$\cos x$偶次幂的积分, 适用半角公式(降幂公式) $\cos^2 x = \frac{1 + \cos 2x}{2}$.

▶答案

$\frac{1}{2}x + \frac{1}{4}\sin 2x + C$

▶详解

【解】

$$\begin{aligned} \int \cos^2 x \mathrm{~d}x &= \int \frac{1 + \cos 2x}{2} \mathrm{~d}x \\ &= \frac{1}{2}x + \frac{1}{4}\sin 2x + C. \end{aligned}$$

第一类换元积分法 (凑微分法)

凑微分的本质是求原函数.

下面是凑微分求不定积分的方法:

$$\begin{aligned} \int f(\varphi(x)) \varphi^{\prime}(x) \mathrm{d} x &= \int f(\varphi(x)) \mathrm{d} \varphi(x) \\ &\xlongequal{\varphi(x)=u} \int f(u) \mathrm{d} u \\ &= F(u)+\mathrm{C} \\ &= F(\varphi(x))+C \end{aligned}$$

填入适当的函数使等式成立:

(1) $\cos x \mathrm{~d} x=\mathrm{d}$ ______ ;

(2) $\frac{1}{x} \mathrm{~d} x=\mathrm{d}$ ______ ;

(3) $\mathrm{e}^{2 x} \mathrm{~d} x=\mathrm{d}$ ______ .

▶答案

(1) $\sin x$

(2) $\ln|x|$

(3) $\frac{1}{2}\mathrm{e}^{2x}$

▶详解

【解】 (1) 因为 $(\sin x)' = \cos x$, 故

$$\cos x \mathrm{~d} x = \mathrm{d}(\sin x).$$

(2) 因为 $(\ln|x|)' = \frac{1}{x}$, 故

$$\frac{1}{x} \mathrm{~d} x = \mathrm{d}(\ln|x|).$$

(如果默认 $x>0$, 填 $\ln x$ 亦可, 但加绝对值更为严谨).

(3) 因为 $\left(\frac{1}{2}\mathrm{e}^{2x}\right)' = \mathrm{e}^{2x}$, 故

$$\mathrm{e}^{2 x} \mathrm{~d} x = \mathrm{d}\left(\frac{1}{2}\mathrm{e}^{2x}\right).$$

求下列各不定积分:

(1) $\displaystyle\int \mathrm{e}^{-4x+5}\mathrm{~d} x$

(2) $\displaystyle\int \frac{\mathrm{d} x}{\sqrt[3]{2-3 x}}$

(3) $\displaystyle\int x\sin(x^{2})\mathrm{~d} x$

(4) $\displaystyle\int\frac{\arctan x}{2(1+x^{2})}\mathrm{~d} x$

(5) $\displaystyle\int\frac{3\ln^{2}x}{x}\mathrm{~d} x$

▶答案

(1) $-\frac{1}{4}\mathrm{e}^{-4x+5} + C$

(2) $-\frac{1}{2} \sqrt[3]{(2-3x)^2} + C$

(3) $-\frac{1}{2}\cos(x^2) + C$

(4) $\frac{1}{4}\arctan^2 x + C$

(5) $\ln^3 x + C$

▶详解

【解】 (1) 凑一次函数的微分:

$$\begin{aligned} \int \mathrm{e}^{-4x+5}\mathrm{~d} x &= -\frac{1}{4} \int \mathrm{e}^{-4x+5} \mathrm{~d}(-4x+5) \\ &= -\frac{1}{4}\mathrm{e}^{-4x+5} + C. \end{aligned}$$

(2) 将根式写成幂的形式, 并凑一次函数的微分:

$$\begin{aligned} \int \frac{\mathrm{d} x}{\sqrt[3]{2-3 x}} &= \int (2-3x)^{-\frac{1}{3}} \mathrm{~d} x \\ &= -\frac{1}{3} \int (2-3x)^{-\frac{1}{3}} \mathrm{~d}(2-3x) \\ &= -\frac{1}{3} \cdot \frac{3}{2} (2-3x)^{\frac{2}{3}} + C \\ &= -\frac{1}{2} \sqrt[3]{(2-3x)^2} + C. \end{aligned}$$

(3) 观察到 $x\mathrm{~d}x$ 是 $x^2$ 导数的一部分, 故凑 $\mathrm{d}(x^2)$:

$$\begin{aligned} \int x\sin(x^{2})\mathrm{~d} x &= \frac{1}{2} \int \sin(x^2) \mathrm{~d}(x^2) \\ &= -\frac{1}{2}\cos(x^2) + C. \end{aligned}$$

(4) 观察到 $\frac{1}{1+x^2}\mathrm{~d}x$ 正是 $\arctan x$ 的微分:

$$\begin{aligned} \int\frac{\arctan x}{2(1+x^{2})}\mathrm{~d} x &= \frac{1}{2} \int \arctan x \mathrm{~d}(\arctan x) \\ &= \frac{1}{2} \cdot \frac{1}{2}(\arctan x)^2 + C \\ &= \frac{1}{4}\arctan^2 x + C. \end{aligned}$$

(5) 观察到 $\frac{1}{x}\mathrm{~d}x$ 是 $\ln x$ 的微分:

$$\begin{aligned} \int\frac{3\ln^{2}x}{x}\mathrm{~d} x &= 3 \int \ln^2 x \mathrm{~d}(\ln x) \\ &= 3 \cdot \frac{1}{3}\ln^3 x + C \\ &= \ln^3 x + C. \end{aligned}$$

[评注]

必须熟练掌握以下对应关系: 看到 $x \mathrm{~d}x$ 想到 $\mathrm{d}(x^2)$, 看到 $\frac{\mathrm{d}x}{1+x^2}$ 想到 $\mathrm{d}(\arctan x)$, 看到 $\frac{\mathrm{d}x}{x}$ 想到 $\mathrm{d}(\ln x)$.

求下列各不定积分:

(1) $\displaystyle\int \frac{\mathrm{d} x}{1-3 x}$;

(2) $\displaystyle\int \frac{3 x^{3}}{1-x^{4}} \mathrm{~d} x$;

(3) $\displaystyle\int \frac{\sin x}{\cos ^{3} x} \mathrm{~d} x$;

(4) $\displaystyle\int\frac{\arcsin ^2x}{\sqrt{1-x^{2}}}\mathrm{~d} x$.

▶答案

(1) $-\frac{1}{3}\ln|1-3x| + C$

(2) $-\frac{3}{4}\ln|1-x^4| + C$

(3) $\frac{1}{2\cos^2 x} + C$

(4) $\frac{1}{3}\arcsin^3 x + C$

▶详解

【解】 (1) 观察分母为一次函数, 直接凑微分:

$$\begin{aligned} \text{原式} &= -\frac{1}{3}\int \frac{\mathrm{d}(1-3x)}{1-3x} \\ &= -\frac{1}{3}\ln|1-3x| + C. \end{aligned}$$

(2) 分子 $x^3$ 是分母中 $x^4$ 导数的一部分, 将其凑入微分号:

$$\begin{aligned} \text{原式} &= -\frac{3}{4}\int \frac{\mathrm{d}(1-x^4)}{1-x^4} \\ &= -\frac{3}{4}\ln|1-x^4| + C. \end{aligned}$$

(3) 提取分子中的 $\sin x$ 凑成 $\cos x$ 的微分:

$$\begin{aligned} \text{原式} &= -\int \frac{\mathrm{d}(\cos x)}{\cos^3 x} \\ &= -\int (\cos x)^{-3} \mathrm{d}(\cos x) \\ &= \frac{1}{2\cos^2 x} + C. \end{aligned}$$

(4) 被积函数中含有 $\arcsin x$ 及其导数结构 $\frac{1}{\sqrt{1-x^2}}$, 直接整体凑微分:

$$\begin{aligned} \text{原式} &= \int \arcsin^2 x \mathrm{~d}(\arcsin x) \\ &= \frac{1}{3}\arcsin^3 x + C. \end{aligned}$$

[评注]

处理含有复合函数的积分时, 优先观察被积函数中是否同时存在某函数及其导数结构.

证明积分公式 $\int \tan x \mathrm{~d} x=-\ln |\cos x|+C$.

▶提示

证明三角函数基本积分公式的关键在于将其转化为正弦与余弦的商, 然后利用凑微分法(第一类换元法)将分子凑成分母的微分.

▶详解

【证明】 根据正切函数的定义, 将被积函数化为正弦与余弦的商:

$$\begin{aligned} \int \tan x \mathrm{~d} x &= \int \frac{\sin x}{\cos x} \mathrm{~d} x \end{aligned}$$

由于 $(\cos x)' = -\sin x$, 即 $\sin x \mathrm{~d} x = -\mathrm{d}(\cos x)$, 将其代入原积分可得

$$\begin{aligned} \int \frac{\sin x}{\cos x} \mathrm{~d} x &= -\int \frac{1}{\cos x} \mathrm{d}(\cos x) \\ &= -\ln |\cos x| + C. \end{aligned}$$

故积分公式 $\int \tan x \mathrm{~d} x=-\ln |\cos x|+C$ 得证.

证明积分公式$\int \cot x \mathrm{~d} x=\ln |\sin x|+C$.

▶提示

将其转化为正弦与余弦的商, 然后利用凑微分法(第一类换元法)将分子凑成分母的微分.

▶详解

【证明】 利用三角函数定义, 将被积函数改写:

$$\begin{aligned} \int \cot x \mathrm{~d}x &= \int \frac{\cos x}{\sin x} \mathrm{~d}x \\ &=\int \frac{1}{\sin x} \mathrm{d}(\sin x) \end{aligned}$$

得:

$$\int \cot x \mathrm{~d}x = \ln |\sin x| + C$$

第二类换元积分法

直接对被积函数中的变量进行换元的方法叫做第二类换元积分法. 去根号是其常见的用途, 下面以根式整体代换为例, 简单介绍第二类换元法 .

若被积函数中含有类似$\sqrt[n]{a x+b}$, 且凑微分无效时, 可令 $\sqrt[n]{a x+b}=t$进行换元.

计算不定积分 $\displaystyle\int \frac{\sin \sqrt{x}}{\sqrt{x}} \mathrm{~d} x$.

▶答案

$-2\cos \sqrt{x} + C$

▶详解

【解】 令 $t = \sqrt{x}$, 则 $x = t^2$, $\mathrm{~d}x = 2t \mathrm{~d}t$.

$$\begin{aligned} \text{原式} &= \int \frac{\sin t}{t} \cdot 2t \mathrm{~d}t \\ &= 2\int \sin t \mathrm{~d}t \\ &= -2\cos t + C \\ &= -2\cos \sqrt{x} + C. \end{aligned}$$

[评注]

也可直接凑微分 , 由于 $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$, 故 $\frac{1}{\sqrt{x}} \mathrm{~d}x = 2\mathrm{d}(\sqrt{x})$ 得

$$\begin{aligned} \int \frac{\sin \sqrt{x}}{\sqrt{x}} \mathrm{~d} x &= 2 \int \sin \sqrt{x} \mathrm{~d}(\sqrt{x}) \\ &= -2\cos \sqrt{x} + C. \end{aligned}$$

计算不定积分$\int x\sqrt{2+x}\mathrm{~d} x.$

▶答案

$\frac{2}{5}(2+x)^{\frac{5}{2}} - \frac{4}{3}(2+x)^{\frac{3}{2}} +C$

▶详解

【解】 令 $t = \sqrt{2+x}$, 则有 $x = t^2 - 2$, 且 $\mathrm{d}x = 2t\mathrm{~d}t$.

代入原积分进行换元, 得

$$\begin{aligned} \int x\sqrt{2+x}\mathrm{~d} x &= \int (t^2 - 2) \cdot t \cdot 2t\mathrm{~d}t \\ &= 2 \int (t^2 - 2) t^2 \mathrm{~d}t \\ &= 2 \int (t^4 - 2t^2 ) \mathrm{~d}t \\ &= \frac{2}{5}t^5 - \frac{4}{3}t^3+ C. \end{aligned}$$

最后将 $t = \sqrt{2+x}$ 代回, 得到

$$\int x\sqrt{2+x}\mathrm{~d} x = \frac{2}{5}(2+x)^{\frac{5}{2}} - \frac{4}{3}(2+x)^{\frac{3}{2}} +C.$$

求不定积分 $\int x \sqrt{1-x} \mathrm{~d}x$.

▶提示

观察到被积函数中含有根式结构, 可直接利用根式整体代换消去无理式.

▶答案

$\frac{2}{5}(1-x)^{\frac{5}{2}} - \frac{2}{3}(1-x)^{\frac{3}{2}} + C$

▶详解

【解】 令 $t = \sqrt{1-x}$, 则 $x = 1 - t^{2}$, $\mathrm{d}x = -2t \mathrm{~d}t$.

$$\begin{aligned} \int x \sqrt{1-x} \mathrm{~d}x &= \int (1 - t^{2}) \cdot t \cdot (-2t) \mathrm{~d}t \\ &= \int (2t^{4} - 2t^{2}) \mathrm{~d}t \\ &= \frac{2}{5}t^{5} - \frac{2}{3}t^{3} + C \\ &= \frac{2}{5}(1-x)^{\frac{5}{2}} - \frac{2}{3}(1-x)^{\frac{3}{2}} + C. \end{aligned}$$

[评注]

本题主要考察根式代换法. 对于含有 $\sqrt{ax+b}$ 的积分, 常用的策略是令 $t = \sqrt{ax+b}$ 以消去根号.

分部积分法

若 $u=u(x), v=v(x)$ 均有连续的导数, 则

$$\int u(x) v^{\prime}(x) \mathrm{d} x=u(x) v(x)-\int u^{\prime}(x) v(x) \mathrm{d} x .$$

即

$$\int u(x) \mathrm{d} v(x)=u(x) v(x)-\int v(x) \mathrm{d} u(x).$$

[评注]

分部积分常常适用于求解两种不同函数相乘时的积分, 要记住口诀: “反对幂三指”,这几种函数相乘时, 谁的次序靠后, 就对谁凑微分.

求下列不定积分:

(1) $\displaystyle\int x \mathrm{e}^ x \mathrm{~d} x$;

(2) $\displaystyle\int x \sin x \mathrm{~d} x$;

(3) $\displaystyle\int x \ln x \mathrm{~d} x$;

(4) $\displaystyle\int \arctan x \mathrm{~d} x$;

(5) $\displaystyle\int \mathrm{e}^{x} \sin x \mathrm{~d} x$.

▶答案

(1) $x\mathrm{e}^x - \mathrm{e}^x + C$

(2) $- x\cos x + \sin x + C$

(3) $\frac{1}{2} x^2\ln x - \frac{1}{4} x^2 + C$

(4) $x \arctan x - \frac{1}{2} \ln(1+x^2) + C$

(5) $\frac{\mathrm{e}^x}{2}(\sin x - \cos x) + C$

▶详解

【解】 (1) 将指数函数凑入微分号, 连续计算得

$$\begin{aligned} \int x \mathrm{e}^x \mathrm{~d} x &= \int x \mathrm{~d}(\mathrm{e}^x) \\ &= x\mathrm{e}^x - \int \mathrm{e}^x \mathrm{~d} x \\ &= x\mathrm{e}^x - \mathrm{e}^x + C. \end{aligned}$$

(2) 将三角函数凑入微分号:

$$\begin{aligned} \int x \sin x \mathrm{~d} x &= - \int x \mathrm{~d}\left(\cos x\right) \\ &= -\left( x\cos x- \int \cos x \mathrm{~d} x \right) \\ &= - x\cos x + \sin x + C. \end{aligned}$$

(3) 遇到对数函数, 将幂函数凑入微分号, 以便对数函数能够求导化简:

$$\begin{aligned} \int x \ln x \mathrm{~d} x &= \frac{1}{2} \int \ln x \mathrm{~d}(x^2) \\ &= \frac{1}{2} x^2\ln x - \frac{1}{2} \int x^2 \mathrm{~d}(\ln x) \\ &= \frac{1}{2} x^2\ln x - \frac{1}{2} \int x \mathrm{~d} x \\ &= \frac{1}{2} x^2\ln x - \frac{1}{4} x^2 + C. \end{aligned}$$

(4) 被积函数只有反三角函数, 视作乘以常数 $1$, 将 $1$ 凑入微分号:

$$\begin{aligned} \int \arctan x \mathrm{~d} x &= \int \arctan x \mathrm{~d}(x) \\ &= x \arctan x - \int x \mathrm{~d}(\arctan x) \\ &= x \arctan x - \int \frac{x}{1+x^2} \mathrm{~d} x \\ &= x \arctan x - \frac{1}{2} \ln(1+x^2) + C. \end{aligned}$$

(5) 指数函数与三角函数相乘,凑哪个函数均可,但需连续两次分部积分, 且这两次凑同类的函数 :

$$\begin{aligned} \int \mathrm{e}^x \sin x \mathrm{~d} x &= \int \sin x \mathrm{~d}(\mathrm{e}^x) \\ &= \mathrm{e}^x \sin x - \int \mathrm{e}^x \cos x \mathrm{~d} x \\ &= \mathrm{e}^x \sin x - \int \cos x \mathrm{~d}(\mathrm{e}^x) \\ &= \mathrm{e}^x \sin x - \left( \mathrm{e}^x \cos x - \int \mathrm{e}^x (-\sin x) \mathrm{~d} x \right) \\ &= \mathrm{e}^x \sin x - \mathrm{e}^x \cos x - \int \mathrm{e}^x \sin x \mathrm{~d} x. \end{aligned}$$

将等式右侧的同类积分项移项至左侧并合并, 除以 $2$ 后加上任意常数, 得

$$\int \mathrm{e}^{x} \sin x \mathrm{~d} x = \frac{\mathrm{e}^x}{2}(\sin x - \cos x) + C.$$

求下列不定积分:

(1) $\int x^2 \ln x \mathrm{~d}x$;

(2) $\int x^2 \sin x \mathrm{~d}x$;

▶答案

(1) $\frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C$

(2) $- x^2 \cos x + 2 x \sin x + 2 \cos x + C$

▶详解

【解】 (1) “幂”与“对”相乘, 凑幂函数$x^2 \mathrm{~d}x=\frac{1}{3}\mathrm{~d}x^3$.

$$\begin{aligned} \int x^2 \ln x \mathrm{~d}x &= \frac{1}{3} \int \ln x \mathrm{~d}(x^3) \\ &= \frac{1}{3} x^3 \ln x - \frac{1}{3} \int x^3 \mathrm{~d}(\ln x) \\ &= \frac{1}{3} x^3 \ln x - \frac{1}{3} \int x^2 \mathrm{~d}x \\ &= \frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C. \end{aligned}$$

[评注]

对于对数函数与多项式相乘的不定积分, 优先将多项式凑微分, 从而利用分部积分法将对数函数求导化简消除.

(2) “幂”与“三”相乘, 凑三角函数的微分, 需使用两次分部积分法

$$\begin{aligned} \int x^2 \sin x \mathrm{~d}x &= - \int x^2 \mathrm{~d}(\cos x) \\ &= - x^2 \cos x + \int \cos x \mathrm{~d}(x^2) \\ &= - x^2 \cos x + 2 \int x \cos x \mathrm{~d}x \\ &= - x^2 \cos x + 2 \int x \mathrm{~d}(\sin x) \\ &= - x^2 \cos x + 2 x \sin x - 2 \int \sin x \mathrm{~d}x \\ &= - x^2 \cos x + 2 x \sin x + 2 \cos x + C. \end{aligned}$$