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Question Number 7030 by Tawakalitu. last updated on 07/Aug/16

∫ ((2x^2 )/(√(1 − x^4 ))) dx

$$\int\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{4}} }}\:{dx}\: \\ $$

Answered by uchechukwu okorie favour last updated on 10/Aug/16

let u=(√(1−x^4 )) (du/dx)=−((2x^3 )/(√(1−x^4 ))) (dx/du)=−((√(1−x^4 ))/(2x)) ⇒∫((2x^3 )/(√(1−x^4 )))dx = ∫((2x^3 )/(√(1−x^4 ))).(dx/du).(du/dx).dx ⇒∫((2x^3 )/(√(1−x^4 )))dx=∫((2x^3 )/(√(1−x^4 ))).−((√(1−x^4 ))/(2x)) =−((√(1−x^4 ))/x)∫(1/(√(1−x^4 ))) =−((√(1−x^4 ))/x)In∣1−x^4 ∣

$${let}\:{u}=\sqrt{\mathrm{1}−{x}^{\mathrm{4}} } \\ $$$$\frac{{du}}{{dx}}=−\frac{\mathrm{2}{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$$$\frac{{dx}}{{du}}=−\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{\mathrm{2}{x}} \\ $$$$\Rightarrow\int\frac{\mathrm{2}{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{dx}\:=\:\int\frac{\mathrm{2}{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}.\frac{{dx}}{{du}}.\frac{{du}}{{dx}}.{dx} \\ $$$$\Rightarrow\int\frac{\mathrm{2}{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{dx}=\int\frac{\mathrm{2}{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}.−\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{\mathrm{2}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{{x}}\int\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}{{x}}{In}\mid\mathrm{1}−{x}^{\mathrm{4}} \mid \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$

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