Tính \(\displaystyle \int\limits_1^2 {({z^2} + 1)\sqrt[3]{{{{(z - 1)}^2}}}} dz\)
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Lời giải:
Báo saiĐặt \(\displaystyle u = \sqrt[3]{{{{(z - 1)}^2}}}\) \(\displaystyle \Rightarrow {u^3} = {\left( {z - 1} \right)^2}\) \(\displaystyle \Rightarrow z = 1 + {u^{\frac{3}{2}}} \Rightarrow dz = \frac{3}{2}{u^{\frac{1}{2}}}du\)
\(\displaystyle \Rightarrow \int\limits_1^2 {({z^2} + 1)\sqrt[3]{{{{(z - 1)}^2}}}} dz\) \(\displaystyle = \int\limits_0^1 {\left[ {{{\left( {1 + {u^{\frac{3}{2}}}} \right)}^2} + 1} \right].u.\frac{3}{2}{u^{\frac{1}{2}}}du} \) \(\displaystyle = \frac{3}{2}\int\limits_0^1 {{u^{\frac{3}{2}}}\left( {2 + 2{u^{\frac{3}{2}}} + {u^3}} \right)du} \)
\(\displaystyle = \frac{3}{2}\int\limits_0^1 {\left( {2{u^{\frac{3}{2}}} + 2{u^3} + {u^{\frac{9}{2}}}} \right)du} \) \(\displaystyle = \frac{3}{2}\left. {\left( {2.\frac{2}{5}{u^{\frac{5}{2}}} + 2.\frac{{{u^4}}}{4} + \frac{2}{{11}}{u^{\frac{{11}}{2}}}} \right)} \right|_0^1\) \(\displaystyle = \frac{3}{2}\left( {\frac{4}{5} + \frac{1}{2} + \frac{2}{{11}}} \right) = \frac{{489}}{{220}}\)