同济高数 第4章 第4-3-23题
📝 题目
23． $\displaystyle{\int} x \ln ^{2} x \mathrm{~d} x$ ．
💡 答案解析
[AI解答]
[AI解答]
我们要求解不定积分
$$
\int x \ln^{2} x \, \mathrm{d}x
$$
使用分部积分法。设
$$
u = \ln^{2} x, \quad \mathrm{d}v = x \, \mathrm{d}x
$$
则
$$
\mathrm{d}u = 2 \ln x \cdot \frac{1}{x} \, \mathrm{d}x = \frac{2 \ln x}{x} \, \mathrm{d}x, \quad v = \frac{x^{2}}{2}
$$
由分部积分公式 $\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u$ 得
$$
\int x \ln^{2} x \, \mathrm{d}x = \frac{x^{2}}{2} \ln^{2} x - \int \frac{x^{2}}{2} \cdot \frac{2 \ln x}{x} \, \mathrm{d}x
$$
化简第二项：
$$
= \frac{x^{2}}{2} \ln^{2} x - \int x \ln x \, \mathrm{d}x
$$
再对 $\int x \ln x \, \mathrm{d}x$ 用分部积分，设
$$
u = \ln x, \quad \mathrm{d}v = x \, \mathrm{d}x
$$
则
$$
\mathrm{d}u = \frac{1}{x} \mathrm{d}x, \quad v = \frac{x^{2}}{2}
$$
于是
$$
\int x \ln x \, \mathrm{d}x = \frac{x^{2}}{2} \ln x - \int \frac{x^{2}}{2} \cdot \frac{1}{x} \, \mathrm{d}x
= \frac{x^{2}}{2} \ln x - \frac{1}{2} \int x \, \mathrm{d}x
$$
$$
= \frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} + C_1
$$
代回原式：
$$
\int x \ln^{2} x \, \mathrm{d}x = \frac{x^{2}}{2} \ln^{2} x - \left( \frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} \right) + C
$$
整理得
$$
\boxed{\frac{x^{2}}{2} \ln^{2} x - \frac{x^{2}}{2} \ln x + \frac{x^{2}}{4} + C}
$$
难度：★★☆☆☆