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Question Number 135145 by ZiYangLee last updated on 10/Mar/21

$$\int_{\mathrm{0}} ^{\mathrm{3}} {x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\:{dx}=? \\$$

Answered by bramlexs22 last updated on 10/Mar/21

$$\int_{\mathrm{0}} ^{\:\mathrm{3}} \:\mathrm{x}^{\mathrm{2}} .\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{9}}\:\mathrm{dx}\:=\:\mathcal{X} \\$$ $$\mathrm{let}\:\mathrm{x}^{\mathrm{2}} +\mathrm{9}\:=\:\phi^{\mathrm{2}} \:\Rightarrow\mathrm{xdx}\:=\:\phi\mathrm{d}\phi\: \\$$ $$\mathcal{X}=\int_{\mathrm{3}} ^{\:\mathrm{3}\sqrt{\mathrm{2}}} \left(\phi^{\mathrm{2}} −\mathrm{9}\right)\phi\:.\left(\phi\mathrm{d}\phi\right) \\$$ $$=\:\int_{\mathrm{0}} ^{\:\mathrm{3}\sqrt{\mathrm{2}}\:} \left(\phi^{\mathrm{4}} −\mathrm{9}\phi\right)\mathrm{d}\phi \\$$ $$=\:\left[\frac{\mathrm{1}}{\mathrm{5}}\phi^{\mathrm{5}} −\mathrm{3}\phi^{\mathrm{2}} \:\right]_{\mathrm{3}} ^{\mathrm{3}\sqrt{\mathrm{2}}} \\$$ $$=\:\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{243}×\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{243}\right)−\mathrm{3}\left(\mathrm{9}\right) \\$$ $$=\frac{\mathrm{243}\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{5}}−\mathrm{27} \\$$

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