📝 题目
例 2 求 $\displaystyle{\int {x}^{k}\ln x\mathrm{\;d}x}$ .
💡 答案与解析
解 先设 $k \neq - 1$ ,则有
$$ \int {x}^{k}\ln x\mathrm{\;d}x = \frac{1}{k + 1}\int \ln x\mathrm{\;d}\left( {x}^{k + 1}\right) $$
$$ = \frac{1}{k + 1}{x}^{k + 1}\ln x - \frac{1}{k + 1}\int {x}^{k + 1}\mathrm{\;d}\left( {\ln x}\right) $$
$$ = \frac{1}{k + 1}{x}^{k + 1}\ln x - \frac{1}{k + 1}\int {x}^{k}\mathrm{\;d}x $$
$$ = \frac{1}{k + 1}{x}^{k + 1}\ln x - \frac{1}{{\left( k + 1\right) }^{2}}{x}^{k + 1} + C. $$
对于 $k = - 1$ 的情形,我们有
$$ \int \frac{\ln x}{x}\mathrm{\;d}x = \int \ln x\mathrm{\;d}\left( {\ln x}\right) = \frac{1}{2}{\left( \ln x\right) }^{2} + C. $$