📝 题目
例 16 求不定积分 $\displaystyle{\int {x}^{2}\sqrt{{x}^{2} + 1}\mathrm{\;d}x}$ .
💡 答案与解析
解法 1
原式 $\displaystyle{= \frac{1}{2}\int \sqrt{{x}^{4} + {x}^{2}}\mathrm{\;d}{x}^{2}}$
$$ = \frac{1}{2}\int \sqrt{{\left( {x}^{2} + \frac{1}{2}\right) }^{2} - {\left( \frac{1}{2}\right) }^{2}}\mathrm{\;d}\left( {{x}^{2} + \frac{1}{2}}\right) $$
$$ \overset{u = {x}^{2} + \frac{1}{2}}{ = }\frac{1}{2}\int \sqrt{{u}^{2} - {\left( \frac{1}{2}\right) }^{2}}\mathrm{\;d}u $$
$$ \overset{\text{ 分部积分 }}{ = }\frac{1}{4}u\sqrt{{u}^{2} - \frac{1}{4}} - \frac{1}{16}\ln \left( {u + \sqrt{{u}^{2} - \frac{1}{4}}}\right) + C $$
$$ = \frac{1}{8}x\left( {2{x}^{2} + 1}\right) \sqrt{{x}^{2} + 1} - \frac{1}{8}\ln \left| {x + \sqrt{{x}^{2} + 1}}\right| + C. $$
解法 2 因为
$$ {\left( {x}^{3}\sqrt{{x}^{2} + 1}\right) }^{\prime } = 3{x}^{2}\sqrt{{x}^{2} + 1} + \frac{{x}^{4}}{\sqrt{{x}^{2} + 1}} $$
$$ = 3{x}^{2}\sqrt{{x}^{2} + 1} + \frac{{x}^{4} - 1 + 1}{\sqrt{{x}^{2} + 1}} $$
所以
$$ = 4{x}^{2}\sqrt{{x}^{2} + 1} - \sqrt{{x}^{2} + 1} + \frac{1}{\sqrt{{x}^{2} + 1}}, $$
$$ {x}^{2}\sqrt{{x}^{2} + 1} = \frac{1}{4}\left\lbrack {{\left( {x}^{3}\sqrt{{x}^{2} + 1}\right) }^{\prime } + \sqrt{{x}^{2} + 1} - \frac{1}{\sqrt{{x}^{2} + 1}}}\right\rbrack , $$
因此
$$ \int {x}^{2}\sqrt{{x}^{2} + 1}\mathrm{\;d}x = \frac{1}{4}\left\lbrack {{x}^{3}\sqrt{{x}^{2} + 1} + \frac{1}{2}\sqrt{{x}^{2} + 1}}\right. $$
$$ \left. {-\frac{1}{2}\ln \left| {x + \sqrt{{x}^{2} + 1}}\right| }\right\rbrack + C. $$
解法 3 因为
$$ \int x\sqrt{{x}^{2} + 1}\mathrm{\;d}x = \frac{1}{2}\int \sqrt{{x}^{2} + 1}\mathrm{\;d}\left( {{x}^{2} + 1}\right) $$
$$ = \frac{1}{3}{\left( {x}^{2} + 1\right) }^{\frac{3}{2}} + {C}_{1}, $$
所以
$$ \int {x}^{2}\sqrt{{x}^{2} + 1}\mathrm{\;d}x = \frac{1}{3}\int x\mathrm{\;d}{\left( {x}^{2} + 1\right) }^{\frac{3}{2}} $$
$$ = \frac{1}{3}x{\left( {x}^{2} + 1\right) }^{\frac{3}{2}} - \frac{1}{3}\int {\left( {x}^{2} + 1\right) }^{\frac{3}{2}}\mathrm{\;d}x, \tag{1.10} $$
又
$$ {x}^{2}\sqrt{{x}^{2} + 1} = \left( {{x}^{2} + 1}\right) \sqrt{{x}^{2} + 1} - \sqrt{{x}^{2} + 1} $$
$$ = {\left( {x}^{2} + 1\right) }^{\frac{3}{2}} - \sqrt{{x}^{2} + 1}\text{ . } \tag{1.11} $$
若设 $I = \int {x}^{2}\sqrt{{x}^{2} + 1}\mathrm{\;d}x,J = \frac{1}{3}\int {\left( {x}^{2} + 1\right) }^{\frac{3}{2}}\mathrm{\;d}x$ ,则由 (1.10) 和 (1.11) 式, 有
$$ \begin{cases} I + J & = \frac{1}{3}x{\left( {x}^{2} + 1\right) }^{\frac{3}{2}}, \\ I - {3J} & = - \int \sqrt{{x}^{2} + 1}\mathrm{\;d}x \\ & = - \frac{1}{2}x\sqrt{{x}^{2} + 1} - \frac{1}{2}\ln \left( {x + \sqrt{{x}^{2} + 1}}\right) + {C}_{2}. \end{cases} $$
由此解得
$$ I = \frac{1}{4}x{\left( {x}^{2} + 1\right) }^{\frac{3}{2}} - \frac{1}{8}x\sqrt{{x}^{2} + 1} $$
$$ - \frac{1}{8}\ln \left( {x + \sqrt{{x}^{2} + 1}}\right) + C. $$