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

$$find{\int }_0^1(x^2+1)\sqrt{\frac{1-x}{1+x}}dx$$

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

$${\int }_0^1\frac{x^2\sqrt{1-x}}{\sqrt{1+x}}dx+{\int }_0^1\sqrt{\frac{1-x}{1+x}}dx$$$${\int }_0^1\frac{x^2(1-x)}{\sqrt{1-x^2}}dx+{\int }_0^1\frac{1-x}{\sqrt{1-x^2}}dx$$$${\int }_0^1\frac{x^2}{\sqrt{1-x^2}}dx-{\int }_0^1\frac{x.x^2}{\sqrt{1-x^2}}dx+{\int }_0^1\frac{dx}{\sqrt{1-x^2}}-{\int }_0^1\frac{xdx}{\sqrt{1-x^2}}$$$$=-{\int }_0^1\frac{1-x^2-1}{\sqrt{1-x^2}}dx-{\int }_0^1\frac{x.x^2}{\sqrt{1-x^2}}dx+{\int }_0^1\frac{dx}{\sqrt{1-x^2}}-{\int }_0^1\frac{xdx}{\sqrt{1-x^2}}$$$$={\int }_0^1\frac{dx}{\sqrt{1-x^2}}-{\int }_0^1\sqrt{1-x^2}dx-{\int }_0^1\frac{x.x^2}{\sqrt{1-x^2}}dx+{\int }_0^1\frac{dx}{\sqrt{1-x^2}}-{\int }_0^1\frac{xdx}{\sqrt{1-x^2}}$$$$t^2=1-x^{2}2tdt=-2xdx$$$$-tdt=xdx$$$$2{\int }_0^1\frac{dx}{\sqrt{1-x^2}}-{\int }_0^1\sqrt{1-x^2}dx+{\int }_1^0\frac{1-t^2}{t}\times tdt+{\int }_1^0\frac{tdt}{t}$$$$nowuseformula$$

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