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Question Number 37893 by abdo mathsup 649 cc last updated on 19/Jun/18

$$calculate{\int }_0^1\sqrt{x+\sqrt{x+1}}dx.$$

Commented by math khazana by abdo last updated on 21/Jun/18

$$let\sqrt{x+\sqrt{x+1}}=t\Rightarrow x+\sqrt{x+1}=t^2\Rightarrow$$$$\sqrt{x+1}=t^2-x\Rightarrow x+1={(t^2-x)}^2\Rightarrow$$$$x=t^4-2{xt}^2+x^2-1\Rightarrow x^2-(2t^2+1)x+t^4-1=0\Rightarrow$$$$\mathrm{\Delta }={(2t^2+1)}^2-4(t^4-1)$$$$=4t^4+4t^2+1-4t^4+4=4t^2+5\Rightarrow$$$$x_1=\frac{2t^2+1+\sqrt{4t^2+5}}{2}$$$$x_2=\frac{2t^2+1-\sqrt{4t^2+5}}{2}withconditiont⩾0andx⩾-1$$$$and{t}^2-x⩾0$$$$t^2-x_2=t^2-\frac{2t^2+1-\sqrt{4t^2+5}}{2}$$$$=\frac{-1+\sqrt{4t^2+5}}{2}⩾0$$$$x_2+1=\frac{2t^2+1-\sqrt{4t^2+5}}{2}+1$$$$=\frac{2t^2+3-\sqrt{4t^2+5}}{2}⩾0sowetake$$$$x=\frac{2t^2+1-\sqrt{4t^2+5}}{2}\Rightarrow \frac{dx}{dt}=2t-\frac{1}{2}\frac{8t}{2\sqrt{4t^2+5}}$$$$=2t-\frac{2t}{\sqrt{4t^2+5}}\Rightarrow$$$${\int }_0^1\sqrt{x+\sqrt{x+1}}dx={\int }_1^{\sqrt{1+\sqrt{2}}}t(2t-\frac{2t}{\sqrt{4t^2+5}})dt$$$$=2{\int }_1^{\sqrt{1+\sqrt{2}}}{t}^{2}dt-2{\int }_1^{\sqrt{1+\sqrt{2}}}\frac{t^2}{\sqrt{4t^2+5}}dt$$$$=2{\left[\frac{t^3}{3}\right]}_1^{\sqrt{1+\sqrt{2}}}-\frac{1}{2}{\int }_1^{\sqrt{1+\sqrt{2}}}\frac{4t^2+5-5}{\sqrt{4t^2+5}}dt$$$$=2((1+\sqrt{2})\sqrt{1+\sqrt{2}}-\frac{1}{3})-\frac{1}{2}{\int }_1^{\sqrt{1+\sqrt{2}}}\sqrt{4t^2+5}dt$$$$+\frac{5}{2}{\int }_1^{\sqrt{1+\sqrt{2}}}\frac{dt}{\sqrt{4t^2+5}}changement$$$$2t=\sqrt{5}sh\left(x\right)give$$$${\int }_1^{\sqrt{1+\sqrt{2}}}\sqrt{4t^2+5}dt={\int }_{argsh\left(\frac{2}{\sqrt{5}}\right)}^{arsh\left(\frac{2}{\sqrt{5}}\sqrt{1+\sqrt{2}}\right)}\sqrt{5}ch\left(x\right)\frac{\sqrt{5}}{2}chxdx$$$$=\frac{5}{2}{\int }_{\alpha }^{\beta }\frac{1+ch\left(2x\right)}{2}dx=\frac{5}{4}(\beta -\alpha )+\frac{5}{8}{\left[sh\left(2x\right)\right]}_{\alpha }^{\beta }$$$$=\frac{5}{4}(\beta -\alpha )+\frac{5}{8}(sh\left(2\beta \right)-sh\left(2\alpha \right))....$$

Answered by MJS last updated on 19/Jun/18

$$\int \sqrt{x+\sqrt{x+1}}dx=$$$$[t=\sqrt{x+1}\to dx=2\sqrt{x+1}dt]$$$$=2\int t\sqrt{t^2+t-1}dt=$$$$=\int (2t+1)\sqrt{t^2+t-1}dt-\int \sqrt{t^2+t-1}dt=$$$$$$$$\int (2t+1)\sqrt{t^2+t-1}dt=$$$$[u=t^2+t-1\to dt=\frac{du}{2t+1}]$$$$=\int \sqrt{u}du=\frac{2}{3}\sqrt{u^3}=\frac{2}{3}\sqrt{{(t^2+t-1)}^3}=$$$$=\frac{2}{3}\sqrt{{(x+\sqrt{x+1})}^3}$$$$$$$$\int \sqrt{t^2+t-1}dt=\int \sqrt{{(t+\frac{1}{2})}^2-\frac{5}{4}}dt=$$$$=\frac{1}{2}\int \sqrt{{(2t+1)}^2-5}dt=$$$$[v=2t+1\to dt=\frac{dv}{2}]$$$$=\frac{1}{4}\int \sqrt{v^2-5}dv=$$$$=\frac{1}{4}\left(\frac{1}{2}(v\sqrt{v^2-5}-5\ln (v+\sqrt{v^2-5}))\right)=$$$$=\frac{1}{8}(2t+1)\sqrt{{(2t+1)}^2-5}-$$$$-\frac{5}{8}\ln (2t+1+\sqrt{{(2t+1)}^2-5})=$$$$=\frac{1}{4}(1+2\sqrt{x+1})\sqrt{x+\sqrt{x+1}}-$$$$-\frac{5}{8}\ln (1+2\sqrt{x+1}+2\sqrt{x+\sqrt{x+1}})$$$$$$$$=\frac{1}{12}\sqrt{x+\sqrt{x+1}}(8x-3+2\sqrt{x+1})+\frac{5}{8}\ln (1+2\sqrt{x+1}+2\sqrt{x+\sqrt{x+1}})+C$$

Commented by prof Abdo imad last updated on 19/Jun/18

$$thankssirMjsyouarereallytalented.$$

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