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Question Number 166301 by LEKOUMA last updated on 18/Feb/22

$$\int \frac{x}{\sqrt{1+x^2+\sqrt{{(1+x^2)}^3}}}dx$$

Answered by MJS_new last updated on 18/Feb/22

$$\int \frac{x}{\sqrt{x^2+1+{(x^2+1)}^{3/2}}}dx=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}dt]$$$$=\frac{\sqrt{2}}{2}\int \frac{t-1}{t^{3/2}}dt=\frac{\sqrt{2}}{2}(2t^{1/2}+\frac{2}{t^{1/2}})=\frac{\sqrt{2}(t+1)}{\sqrt{t}}=$$$$...$$$$=2\sqrt{1+\sqrt{x^2+1}}+C$$

Commented by cortano1 last updated on 18/Feb/22

$$\mathrm{Euler}\mathrm{subtitution}$$

Commented by peter frank last updated on 19/Feb/22

$$\mathrm{thank}\mathrm{you}$$

Answered by cortano1 last updated on 18/Feb/22

$$\int \frac{1}{\sqrt{1+{\mathrm{x}}^2+\sqrt{{(1+{\mathrm{x}}^2)}^3}}}\left(\mathrm{x}\mathrm{dx}\right)$$$$\mathrm{let}\mathrm{y}=\sqrt{1+{\mathrm{x}}^2}\Rightarrow {\mathrm{y}}^2=1+{\mathrm{x}}^2$$$$\Rightarrow \mathrm{y}\mathrm{dy}=\mathrm{x}\mathrm{dx}$$$$\mathrm{I}=\int \frac{\mathrm{y}\mathrm{dy}}{\sqrt{{\mathrm{y}}^2+{\mathrm{y}}^3}}=\int \frac{\mathrm{dy}}{\sqrt{1+\mathrm{y}}}$$$$\mathrm{I}=\int {(1+\mathrm{y})}^{-\frac{1}{2}}\mathrm{dy}=2\sqrt{1+\mathrm{y}}+\mathrm{c}$$$$\mathrm{I}=2\sqrt{1+\sqrt{1+{\mathrm{x}}^2}}+\mathrm{c}$$

Commented by LEKOUMA last updated on 18/Feb/22

$$Thanks$$

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