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Question Number 60494 by Mr X pcx last updated on 21/May/19

$$find\int \sqrt{\sqrt{2+x^2}-x}dx$$

Commented by maxmathsup by imad last updated on 22/May/19

$$letusethechang.\sqrt{2+x^2}-x=t\Rightarrow \sqrt{2+x^2}=x+t\Rightarrow 2+x^2=x^2+2xt+t^2\Rightarrow$$$$2xt+t^2=2\Rightarrow 2tx=2-t^2\Rightarrow x=\frac{2-t^2}{2t}=\frac{1}{t}-\frac{t}{2}\Rightarrow dx=(-\frac{1}{t^2}-\frac{1}{2})dt\Rightarrow$$$$\int \sqrt{\sqrt{2+x^2}-x}dx=\int \sqrt{t}(-\frac{1}{t^2}-\frac{1}{2})dt$$$$=-\int \frac{\sqrt{t}}{t^2}dt-\frac{1}{2}\int \sqrt{t}dt{=}_{\sqrt{t}=u}-\int \frac{u}{u^4}\left(2u\right)du-\frac{1}{2}\int u\left(2u\right)du$$$$=-2\int \frac{du}{u^2}-\int u^{2}du=\frac{2}{u}-\frac{1}{3}{u}^3+c=\frac{2}{\sqrt{t}}-\frac{1}{3}{\left(\sqrt{t}\right)}^3+c$$$$=\frac{2}{\sqrt{\sqrt{2+x^2-x}}}-\frac{1}{3}{\left(\sqrt{\sqrt{2+x^2}-x}\right)}^3+c$$

Answered by MJS last updated on 21/May/19

$$\int \sqrt{\sqrt{x^2+2}-x}dx=$$$$[t=\sqrt{\sqrt{x^2+2}-x}\to dx=-\frac{2\sqrt{x^2+2}}{\sqrt{\sqrt{x^2+2}-x}}dt]$$$$[\Rightarrow x=\frac{2-t^4}{2t^2}\Rightarrow dx=-\frac{t^4+2}{t^3}dt]$$$$=-\int \frac{t^4+2}{t^2}dt=-\int t^{2}dt-2\int \frac{dt}{t^2}=-\frac{t^3}{3}+\frac{2}{t}=$$$$=\frac{6-t^4}{3t}=\frac{2}{3}(x\sqrt{\sqrt{x^2+2}-x}+\frac{2}{\sqrt{\sqrt{x^2+2}-x}})+C$$

Commented by maxmathsup by imad last updated on 22/May/19

$$thankssirmjs.$$

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