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Question Number 29441 by prof Abdo imad last updated on 08/Feb/18

find ∫  (x^2 /((2−x^2 )(√(1−x^2 ))))dx

$${find}\:\int\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$

Commented by prof Abdo imad last updated on 07/Mar/18

I =−∫ ((2−x^2  −2)/((2−x^2 )(√(1−x^2 ))))dx= −∫  (dx/(√(1−x^2 ))) +2∫   (dx/((2−x^2 )(√(1−x^2 ))))  =−arcsinx +2 ∫    (dx/((2−x^2 )(√(1−x^2 )))) the ch.x=sint⇒  ∫     (dx/((2−x^2 )(√(1−x^2 )))) =∫    ((cost dt)/((2−sin^2 t)cost))  = ∫    (dt/(1+cos^2 t)) =∫    (dt/(1+((1+cos(2t))/2)))  = ∫     ((2dt)/(3+cos(2t))) and tbe type of this integral  is calculated....

$${I}\:=−\int\:\frac{\mathrm{2}−{x}^{\mathrm{2}} \:−\mathrm{2}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\:−\int\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+\mathrm{2}\int\:\:\:\frac{{dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=−{arcsinx}\:+\mathrm{2}\:\int\:\:\:\:\frac{{dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{the}\:{ch}.{x}={sint}\Rightarrow \\ $$$$\int\:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\int\:\:\:\:\frac{{cost}\:{dt}}{\left(\mathrm{2}−{sin}^{\mathrm{2}} {t}\right){cost}} \\ $$$$=\:\int\:\:\:\:\frac{{dt}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:=\int\:\:\:\:\frac{{dt}}{\mathrm{1}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}} \\ $$$$=\:\int\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}\:{and}\:{tbe}\:{type}\:{of}\:{this}\:{integral} \\ $$$${is}\:{calculated}.... \\ $$

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