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Integration Questions

Question Number 67835 by mind is power last updated on 01/Sep/19

$$\int_{\mathrm{0}} ^{\mathrm{2}} {x}\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx} \\$$

Answered by MJS last updated on 01/Sep/19

$$\int{x}\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} = \\$$ $$\:\:\:\:\:\left[{t}=\frac{\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} }{{x}}\:\rightarrow\:{dx}=−\frac{{x}^{\mathrm{2}} \left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{8}}{dt}\right] \\$$ $$=−\mathrm{8}\int\frac{{t}^{\mathrm{3}} }{\left({t}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\$$ $$\:\:\:\:\:\mathrm{now}\:\mathrm{decompose}\:\mathrm{and}\:\mathrm{solve};\:\mathrm{I}\:\mathrm{get} \\$$ $$=\frac{\mathrm{8}{t}}{\mathrm{3}\left({t}^{\mathrm{3}} +\mathrm{1}\right)}+\frac{\mathrm{4}}{\mathrm{9}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}}}\:= \\$$ $$=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} \left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} +\frac{\mathrm{4}}{\mathrm{9}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} −{x}\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{2}/\mathrm{3}} }{\left({x}+\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \right)^{\mathrm{2}} }\:−\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\mathrm{2}\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} −{x}}{\sqrt{\mathrm{3}}{x}}\:+{Cg} \\$$ $$\\$$ $$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{x}\left(\mathrm{8}−{x}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} =\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\mathrm{27}}\pi \\$$

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