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# 含定积分的𝘶-代入法

${\int }_{1}^{2}2x\left({x}^{2}+1{\right)}^{3}\phantom{\rule{0.167em}{0ex}}dx={\int }_{1}^{2}\left(u{\right)}^{3}\phantom{\rule{0.167em}{0ex}}du$

• 下限：$\left(1{\right)}^{2}+1=2$
• 上限：$\left(2{\right)}^{2}+1=5$

${\int }_{1}^{2}2x\left({x}^{2}+1{\right)}^{3}\phantom{\rule{0.167em}{0ex}}dx={\int }_{2}^{5}\left(u{\right)}^{3}\phantom{\rule{0.167em}{0ex}}du$

$\begin{array}{rl}{\int }_{2}^{5}{u}^{3}\phantom{\rule{0.167em}{0ex}}du& ={\left[\frac{{u}^{4}}{4}\right]}_{2}^{5}\\ \\ & =\frac{{5}^{4}}{4}-\frac{{2}^{4}}{4}\\ \\ & =152.25\end{array}$

${\int }_{1}^{5}\left(2x+1\right)\left({x}^{2}+x{\right)}^{3}dx={\int }_{1}^{5}{u}^{3}du$

$\begin{array}{rl}{\int }_{1}^{5}{u}^{3}du& ={\left[\frac{{u}^{4}}{4}\right]}_{1}^{5}\\ \\ & =\frac{{5}^{4}}{4}-\frac{{1}^{4}}{4}\\ \\ & =156\end{array}$

${\int }_{1}^{2}15{x}^{2}\left({x}^{3}-7{\right)}^{4}\phantom{\rule{0.167em}{0ex}}dx=\phantom{\rule{0.167em}{0ex}}?$