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Question Number 83713 by mhmd last updated on 05/Mar/20

$$\int \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x}}}}pleassirhelpme$$

Commented by niroj last updated on 05/Mar/20

$$$$$$\int \frac{1}{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}\mathrm{dx}$$$$=\int \frac{1}{\frac{1}{2}+\frac{\sqrt{4\mathrm{x}+1}}{2}}\mathrm{dx}=\int \frac{2}{1+\sqrt{4\mathrm{x}+1}}\mathrm{dx}$$$$\mathrm{put}\sqrt{4\mathrm{x}+1}=\mathrm{t}$$$$4\mathrm{x}+1={\mathrm{t}}^2$$$$4\mathrm{dx}=2\mathrm{tdt}$$$$\mathrm{dx}=\frac{\mathrm{t}}{2}\mathrm{dt}$$$$\int \frac{2}{1+\mathrm{t}}.\frac{\mathrm{t}}{2}\mathrm{dt}=\int \frac{\mathrm{t}}{\mathrm{t}+1}\mathrm{dt}$$$$=\int \frac{\mathrm{t}+1-1}{\mathrm{t}+1}\mathrm{dt}=\int \mathrm{dt}-\int \frac{1}{\mathrm{t}+1}\mathrm{dt}$$$$=\mathrm{t}-\log (\mathrm{t}+1)+\mathrm{C}$$$$=\sqrt{4\mathrm{x}+1}-\log (\sqrt{4\mathrm{x}+1}+1)+\mathrm{C}//.$$

Commented by MJS last updated on 05/Mar/20

$$\mathrm{how}\sqrt{x+\sqrt{x+\sqrt{x}}}=\frac{1}{2}(1+\sqrt{4x+1})?$$

Commented by mhmd last updated on 05/Mar/20

$$imtoodontknowthis?$$

Commented by jagoll last updated on 05/Mar/20

$$???$$

Commented by mhmd last updated on 05/Mar/20

$$howthissircanihelpme?$$

Commented by MJS last updated on 05/Mar/20

$$\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}\sqrt{x+\sqrt{x+\sqrt{x+...}}}\mathrm{to}\mathrm{infinity}:$$$$y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$$$$y^2=x+y$$$$\Rightarrow y=\frac{1}{2}+\frac{1}{2}\sqrt{4x+1}$$$$\mathrm{but}\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}\sqrt{x+\sqrt{x+\sqrt{x}}}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{afraid}\mathrm{we}\mathrm{cannot}$$$$\mathrm{solve}\mathrm{the}\mathrm{integral}$$

Commented by jagoll last updated on 05/Mar/20

$$\mathrm{but}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}\mathrm{not}\mathrm{infinity}$$

Commented by MJS last updated on 05/Mar/20

$$\mathrm{yes}.\Rightarrow \mathrm{I}\mathrm{cannot}\mathrm{solve}\mathrm{it}$$

Answered by msup trace by abdo last updated on 05/Mar/20

$$thisintegralisnotsolvable$$$$bigdifficulltyincalculus...$$

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