📝 题目
9． $\displaystyle{\int} x^{2} \arctan x \mathrm{~d} x$ ．
💡 答案与解析
[AI解答]
[AI解答] 我们要求不定积分：
$$ \displaystyle{\int} x^{2} \arctan x \, \mathrm{d}x $$
**步骤1：分部积分法** 令 $$ u = \arctan x, \quad \mathrm{d}v = x^{2} \mathrm{d}x $$ 则 $$ \mathrm{d}u = \frac{1}{1+x^{2}} \mathrm{d}x, \quad v = \frac{x^{3}}{3} $$
由分部积分公式 $$ \int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u $$ 得 $$ \int x^{2} \arctan x \, \mathrm{d}x = \frac{x^{3}}{3} \arctan x - \int \frac{x^{3}}{3} \cdot \frac{1}{1+x^{2}} \, \mathrm{d}x $$
**步骤2：化简剩余积分** $$ \int \frac{x^{3}}{1+x^{2}} \, \mathrm{d}x $$ 做多项式除法： $$ \frac{x^{3}}{1+x^{2}} = x - \frac{x}{1+x^{2}} $$ 所以 $$ \int \frac{x^{3}}{1+x^{2}} \mathrm{d}x = \int x \, \mathrm{d}x - \int \frac{x}{1+x^{2}} \, \mathrm{d}x $$
**步骤3：分别积分** $$ \int x \, \mathrm{d}x = \frac{x^{2}}{2} $$ 对于第二项，令 $ t = 1+x^{2} $，则 $\mathrm{d}t = 2x \, \mathrm{d}x$，于是 $$ \int \frac{x}{1+x^{2}} \mathrm{d}x = \frac{1}{2} \int \frac{1}{t} \mathrm{d}t = \frac{1}{2} \ln|1+x^{2}| + C $$ 因此 $$ \int \frac{x^{3}}{1+x^{2}} \mathrm{d}x = \frac{x^{2}}{2} - \frac{1}{2} \ln(1+x^{2}) + C $$
**步骤4：代回原式** $$ \int x^{2} \arctan x \, \mathrm{d}x = \frac{x^{3}}{3} \arctan x - \frac{1}{3} \left( \frac{x^{2}}{2} - \frac{1}{2} \ln(1+x^{2}) \right) + C $$ 化简得 $$ \int x^{2} \arctan x \, \mathrm{d}x = \frac{x^{3}}{3} \arctan x - \frac{x^{2}}{6} + \frac{1}{6} \ln(1+x^{2}) + C $$
最终结果为： $$ \boxed{\displaystyle \frac{x^{3}}{3} \arctan x - \frac{x^{2}}{6} + \frac{1}{6} \ln(1+x^{2}) + C} $$
难度：★★☆☆☆