\[\int{\left( {{x}^{6}}+{{x}^{3}} \right){{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}dx=}\int{\left( {{x}^{6}}+2{{x}^{3}}-{{x}^{3}} \right){{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}\,dx=}\int{{{x}^{3}}{{\left( {{x}^{3}}+2 \right)}^{\frac{4}{3}}}\,dx}-\int{{{x}^{3}}{{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}\,dx}\]
e integrando per parti il primo dei due integrali si ha la seguente uguaglianza: \[\int{\left( {{x}^{6}}+{{x}^{3}} \right){{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}dx=}\frac{1}{4}{{x}^{4}}{{\left( {{x}^{3}}+2 \right)}^{\frac{4}{3}}}-\int{{{x}^{6}}{{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}\,dx}-\int{{{x}^{3}}{{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}\,dx}\] da cui: \[\int{\left( {{x}^{6}}+{{x}^{3}} \right){{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}dx=}\frac{1}{4}{{x}^{4}}{{\left( {{x}^{3}}+2 \right)}^{\frac{4}{3}}}-\int{\left( {{x}^{6}}+{{x}^{3}} \right){{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}\,dx}\to \]\[\to 2\int{\left( {{x}^{6}}+{{x}^{3}} \right){{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}dx=}\frac{1}{4}{{x}^{4}}{{\left( {{x}^{3}}+2 \right)}^{\frac{4}{3}}}\to \]\[\to \int{\left( {{x}^{6}}+{{x}^{3}} \right){{\left( {{x}^{3}}+2 \right)}^{\frac{1}{3}}}dx}=\frac{{{x}^{4}}{{\left( {{x}^{3}}+2 \right)}^{\frac{4}{3}}}}{8}+c\quad .\] Massimo BergaminiUn integrale
Ettore propone il calcolo del seguente integrale indefinito:
\[\int{\left( {{x}^{6}}+{{x}^{3}} \right)\sqrt[3]{{{x}^{3}}+2}\,dx}\quad .\]