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Question Number 56700 by maxmathsup by imad last updated on 21/Mar/19

$$find\int \sqrt{x-2\sqrt{x}+3}dx$$

Commented by kaivan.ahmadi last updated on 22/Mar/19

Commented by maxmathsup by imad last updated on 23/Mar/19

$$I=\int \sqrt{{(\sqrt{x}-1)}^2+2}dxchangement\sqrt{x}-1=\sqrt{2}sh\left(t\right)give\sqrt{x}=1+\sqrt{2}sht\Rightarrow$$$$x={(1+\sqrt{2}sh\left(t\right))}^2\Rightarrow$$$$I=\int \sqrt{2{sh}^{2}t+2}(2(1+\sqrt{2}sh\left(t\right))\sqrt{2}chtdt$$$$=4\int {ch}^2\left(t\right)(1+\sqrt{2}sh\left(t\right))dt=4\int {ch}^{2}tdt+4\sqrt{2}\int sh\left(t\right){ch}^{2}tdt$$$$=4\int \frac{1+ch\left(2t\right)}{2}dt+4\sqrt{2}\frac{1}{3}{ch}^3\left(t\right)=2t+2\int ch\left(2t\right)dt+\frac{4\sqrt{2}}{3}{ch}^3\left(t\right)$$$$=2t+sh\left(2t\right)+\frac{4\sqrt{2}}{3}{ch}^3\left(t\right)butt=argsh\left(\frac{\sqrt{x}-1}{2}\right)=ln(\frac{\sqrt{x}-1}{2}+\sqrt{1+{\left(\frac{\sqrt{x}-1}{2}\right)}^2})$$$$I=2argsh\left(\frac{\sqrt{x}-1}{2}\right)+sh\left(2argsh\left(\frac{\sqrt{x}-1}{2}\right)\right)+\frac{4\sqrt{2}}{3}{ch}^3\left(argsh\left(\frac{\sqrt{x}-1}{2}\right)\right)+c$$

Answered by kaivan.ahmadi last updated on 22/Mar/19

$$\int \sqrt{{(\sqrt{x}-1)}^2+2}dx=$$$$t=\sqrt{x}-1\Rightarrow dt=\frac{dx}{2\sqrt{x}}=\frac{dx}{2t+2}\Rightarrow dx=(2t+2)dt$$$$\int \sqrt{t^2+2}(2t+2)dt=\int 2t\sqrt{t^2+2}dt+\int 2\sqrt{t^2+2}dt=I+J$$$$solveI$$$$u=t^2+2\Rightarrow du=2tdt$$$$\int \sqrt{u}du=\frac{2}{3}{u}^{\frac{3}{2}}+C_1=\frac{2}{3}{(t^2+2)}^{\frac{3}{2}}+C_1=$$$${({(\sqrt{x}-1)}^2+2)}^{\frac{3}{2}}+C_1$$$$solveJ$$$$t=\sqrt{2}tg\theta \Rightarrow dt=\sqrt{2}(1+{tg}^2\theta )d\theta$$$$andsotg\theta =\frac{t}{\sqrt{2}}\Rightarrow sec\theta =\sqrt{1+\frac{t^2}{2}}$$$$\int 2\sqrt{2(1+{tg}^2\theta )}.\sqrt{2}(1+{tg}^2\theta )d\theta =$$$$\int 4(1+{tg}^2\theta )\sqrt{1+{tg}^2\theta }d\theta =\int 4\frac{d\theta }{{cos}^3\theta }=$$$$4\int {sec}^3\theta d\theta =4(\frac{1}{2}sec\theta tg\theta +\frac{1}{2}(sec\theta +tg\theta ))+C_2=$$$$2sec\theta tg\theta +2(sec\theta +tg\theta )+C_2=$$$$2\sqrt{1+\frac{t^2}{2}}.\frac{t}{\sqrt{2}}+2(\sqrt{1+\frac{t^2}{2}}+\frac{t}{\sqrt{2}})+C_2$$$$$$

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