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Question Number 42803 by maxmathsup by imad last updated on 02/Sep/18

find ∫_0 ^1 (x^2 +1)(√((1−x)/(1+x)))dx

$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx}\: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18

∫_0 ^1 ((x^2 (√(1−x)))/(√(1+x)))dx+∫_0 ^1 (√(((1−x)/(1+x)) )) dx ∫_0 ^1 ((x^2 (1−x))/(√(1−x^2 )))dx+∫_0 ^1 ((1−x)/(√(1−x^2 )))dx ∫_0 ^1 (x^2 /(√(1−x^2 )))dx−∫_0 ^1 ((x.x^2 )/(√(1−x^2 )))dx+∫_0 ^1 (dx/(√(1−x^2 )))−∫_0 ^1 ((xdx)/(√(1−x^2 ))) =−∫_0 ^1 ((1−x^2 −1)/(√(1−x^2 )))dx−∫_0 ^1 ((x.x^2 )/(√(1−x^2 )))dx+∫_0 ^1 (dx/(√(1−x^2 )))−∫_0 ^1 ((xdx)/(√(1−x^2 ))) =∫_0 ^1 (dx/(√(1−x^2 )))−∫_0 ^1 (√(1−x^2 )) dx−∫_0 ^1 ((x.x^2 )/(√(1−x^2 )))dx+∫_0 ^1 (dx/(√(1−x^2 )))−∫_0 ^1 ((xdx)/(√(1−x^2 ))) t^2 =1−x^2 2tdt=−2xdx −tdt=xdx 2∫_0 ^1 (dx/(√(1−x^2 )))−∫_0 ^1 (√(1−x^2 )) dx+∫_1 ^0 ((1−t^2 )/t)×tdt+∫_1 ^0 ((tdt)/t) now use formula

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}}}{\sqrt{\mathrm{1}+{x}}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}.{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xdx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}.{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xdx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}.{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xdx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${t}^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} \:\:\:\:\mathrm{2}{tdt}=−\mathrm{2}{xdx} \\ $$$$−{tdt}={xdx} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}+\int_{\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}}×{tdt}+\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{tdt}}{{t}} \\ $$$${now}\:{use}\:{formula} \\ $$

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