# Question and Answers Forum

Integration Questions

Question Number 131147 by EDWIN88 last updated on 02/Feb/21

$$\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\sqrt{{x}−{x}^{\mathrm{3}} }}\:{dx}\:=?\: \\$$

Commented by EDWIN88 last updated on 02/Feb/21

$$\:\sqrt{{x}−{x}^{\mathrm{3}} }\:=\:{w}−\mathrm{1}\:\Rightarrow{x}−{x}^{\mathrm{3}} =\left({w}−\mathrm{1}\right)^{\mathrm{2}} \\$$$$\:\left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} \right){dx}=\mathrm{2}\left({w}−\mathrm{1}\right){dw}\: \\$$$${I}=−\int_{\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{2}{w}−\mathrm{1}}{{w}}\:{dw}\:=\:\int_{\mathrm{1}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{{w}}−\mathrm{2}\right){dw} \\$$$$=\:\left[\:\mathrm{ln}\:\mid{w}\mid−\mathrm{2}{w}\:\right]_{\mathrm{1}} ^{\mathrm{1}} \\$$$$=\:−\mathrm{2}\:−\left(−\mathrm{2}\right)\:=\:\mathrm{0} \\$$

Answered by Ar Brandon last updated on 02/Feb/21

$$\mathrm{Let}\:\mathrm{u}=\mathrm{x}−\mathrm{x}^{\mathrm{3}} \:\mathrm{then}\:\mathrm{du}=−\left(\mathrm{3x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{dx} \\$$

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