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 $$ \boxed{\frac{x^{2}}{2} \ln^{2} x - \frac{x^{2}}{2} \ln x + \frac{x^{2}}{4} + C} $$
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