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Question Number 129251 by bemath last updated on 14/Jan/21

Commented by bemath last updated on 14/Jan/21

$$\mathrm{thank}\mathrm{you}\mathrm{all}\mathrm{sirs}$$

Answered by liberty last updated on 14/Jan/21

$$let1+x^3=q\Rightarrow x^{2}dx=\frac{1}{3}dq$$$$\mathcal{G}=\int (q-1){\left(q\right)}^{\frac{2}{3}}\left(\frac{1}{3}dq\right)$$$$\mathcal{G}=\frac{1}{3}\int (q^{\frac{5}{3}}-q^{\frac{2}{3}})dq=\frac{1}{3}(\frac{3}{8}{q}^{\frac{8}{3}}-\frac{3}{5}{q}^{\frac{5}{3}})+C$$$$\underset{―}{\mathcal{G}=\left(\frac{5{\mathit{x}}^3-3}{40}\right)\sqrt[3]{{(1+{\mathit{x}}^3)}^5}+\mathit{C}}$$

Answered by Lordose last updated on 14/Jan/21

$$\mathrm{G}=\int {\mathrm{x}}^5\sqrt[3]{{(1+{\mathrm{x}}^3)}^2}\mathrm{dx}=\int {\mathrm{x}}^5\sqrt[3]{(1+{2\mathrm{x}}^3+{\mathrm{x}}^6)}\mathrm{dx}$$$$\mathrm{u}={\mathrm{x}}^6$$$$\mathrm{G}=\frac{1}{6}\int \sqrt[3]{1+2\sqrt{\mathrm{u}}+\mathrm{u}}\mathrm{du}\stackrel{\mathrm{y}=\sqrt{\mathrm{u}}}{=}\frac{1}{3}\int \mathrm{y}\sqrt[3]{1+2\mathrm{y}+{\mathrm{y}}^2}$$$$\mathrm{G}=\frac{1}{3}\int \mathrm{y}{(1+\mathrm{y})}^{\frac{3}{2}}\mathrm{du}\stackrel{\mathrm{s}=1+\mathrm{y}}{=}\frac{1}{3}\int (\mathrm{s}-1){\mathrm{s}}^{\frac{3}{2}}\mathrm{ds}$$$$\mathrm{G}=\frac{1}{3}\int {\mathrm{s}}^{\frac{5}{2}}\mathrm{ds}-\frac{1}{3}\int {\mathrm{s}}^{\frac{3}{2}}\mathrm{ds}$$$$\mathrm{G}=\frac{{2\mathrm{s}}^{\frac{7}{2}}}{21}-\frac{2}{15}{\mathrm{s}}^{\frac{5}{2}}+\mathrm{C}$$$$\mathrm{s}=(1+\mathrm{y})$$$$\mathrm{y}=\sqrt{\mathrm{u}}$$$$\mathrm{u}={\mathrm{x}}^6$$$$$$

Answered by Olaf last updated on 14/Jan/21

$$\mathcal{G}=\int x^5\sqrt[3]{{(1+x^3)}^2}dx$$$$\mathrm{Let}x={\sinh }^{\frac{2}{3}}u$$$$\mathcal{G}=\int {\sinh }^{\frac{10}{3}}u\sqrt[3]{{(1+{\sinh }^{2}u)}^2(}\frac{2}{3}{\sinh }^{-\frac{1}{3}}u\cosh u)du$$$$\mathcal{G}=\frac{2}{3}\int {\sinh }^{3}u{\cosh }^{\frac{7}{3}}udu$$$$\mathcal{G}=\frac{2}{3}\int \sinh u({\cosh }^{2}u-1){\cosh }^{\frac{7}{3}}udu$$$$\mathcal{G}=\frac{2}{3}\int \sinh u{\cosh }^{\frac{13}{3}}udu-\frac{2}{3}\int \sinh u{\cosh }^{\frac{7}{3}}udu$$$$\mathcal{G}=\frac{2}{3}\times \frac{3}{16}{\cosh }^{\frac{16}{3}}u-\frac{2}{3}\times \frac{3}{10}{\cosh }^{\frac{10}{3}}u+\mathrm{C}$$$$\mathcal{G}=\frac{1}{8}{\cosh }^{\frac{16}{3}}u-\frac{1}{5}{\cosh }^{\frac{10}{3}}u+\mathrm{C}$$$$\mathcal{G}=\frac{1}{8}{\left({\cosh }^{2}u\right)}^{\frac{8}{3}}-\frac{1}{5}{\left({\cosh }^{2}u\right)}^{\frac{5}{3}}+\mathrm{C}$$$$\mathcal{G}=\frac{1}{8}{(1+{\sinh }^{2}u)}^{\frac{8}{3}}-\frac{1}{5}{(1+{\sinh }^{2}u)}^{\frac{5}{3}}+\mathrm{C}$$$$\mathcal{G}=\frac{1}{8}{(1+x^3)}^{\frac{8}{3}}-\frac{1}{5}{(1+x^3)}^{\frac{5}{3}}+\mathrm{C}$$

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