📝 题目
例 14 求不定积分 $\displaystyle \int \frac{x{\mathrm{e}}^{\arctan x}}{{\left( 1 + {x}^{2}\right) }^{\frac{3}{2}}}\mathrm{\;d}x$ .
💡 答案与解析
解法 1
$$ \int \frac{x{\mathrm{e}}^{\arctan x}}{{\left( 1 + {x}^{2}\right) }^{\frac{3}{2}}}\mathrm{\;d}x\overset{x = \tan t}{ = }\int \frac{{\mathrm{e}}^{t}\tan t}{{\left( {\tan }^{2}t + 1\right) }^{\frac{3}{2}}}{\sec }^{2}t\mathrm{\;d}t = \int {\mathrm{e}}^{t}\sin t\mathrm{\;d}t, $$
$$ I\overset{\text{ 定义 }}{ = }\int {\mathrm{e}}^{t}\sin t\mathrm{\;d}t = {\mathrm{e}}^{t}\sin t - \int {\mathrm{e}}^{t}\cos t\mathrm{\;d}t $$
$$ = {\mathrm{e}}^{t}\sin t - J \Rightarrow I + J = {\mathrm{e}}^{t}\sin t, $$
$$ J\frac{\text{ 定义 }}{}\int {\mathrm{e}}^{t}\cos t\mathrm{\;d}t = {\mathrm{e}}^{t}\cos t + \int {\mathrm{e}}^{t}\sin t\mathrm{\;d}t $$
$$ = {\mathrm{e}}^{t}\cos t + I \Rightarrow I - J = - {\mathrm{e}}^{t}\cos t. $$
因此 $I = \frac{1}{2}{\mathrm{e}}^{t}\left( {\sin t - \cos t}\right) + C$ ,
$$ \int \frac{x{\mathrm{e}}^{\arctan x}}{{\left( 1 + {x}^{2}\right) }^{\frac{3}{2}}}\mathrm{\;d}x = I = \frac{1}{2}{\mathrm{e}}^{\arctan x}\left( {\frac{x}{\sqrt{1 + {x}^{2}}} - \frac{1}{\sqrt{1 + {x}^{2}}}}\right) + C $$
$$ = \frac{\left( {x - 1}\right) {\mathrm{e}}^{\arctan x}}{2\sqrt{1 + {x}^{2}}} + C. $$
解法 2
$$ I = \int \frac{x{\mathrm{e}}^{\arctan x}}{{\left( 1 + {x}^{2}\right) }^{\frac{3}{2}}}\mathrm{\;d}x = \int \frac{x}{\sqrt{1 + {x}^{2}}}\mathrm{\;d}{\mathrm{e}}^{\arctan x} $$
$$ \overset{\text{ 分部积分 }}{ = }\frac{x{\mathrm{e}}^{\arctan x}}{\sqrt{1 + {x}^{2}}} - \int \frac{{\mathrm{e}}^{\arctan x}}{{\left( 1 + {x}^{2}\right) }^{\frac{3}{2}}}\mathrm{\;d}x $$
$$ = \frac{x{\mathrm{e}}^{\arctan x}}{\sqrt{1 + {x}^{2}}} - \int \frac{1}{\sqrt{1 + {x}^{2}}}\mathrm{\;d}{\mathrm{e}}^{\arctan x} $$
$$ = \frac{x{\mathrm{e}}^{\arctan x}}{\sqrt{1 + {x}^{2}}} - \frac{{\mathrm{e}}^{\arctan x}}{\sqrt{1 + {x}^{2}}} - I. $$
由此推出 $I = \frac{\left( {x - 1}\right) {\mathrm{e}}^{\arctan x}}{2\sqrt{1 + {x}^{2}}} + C$ .