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Question Number 123234 by benjo_mathlover last updated on 24/Nov/20

$$\int \sqrt{x^2-4x+5}dx$$

Commented by liberty last updated on 24/Nov/20

Answered by MJS_new last updated on 24/Nov/20

$$\int \sqrt{x^2-4x+5}dx=$$$$[t=x-2+\sqrt{x^2-4x+5}\to dx=\frac{\sqrt{x^2-4x+5}}{x-2+\sqrt{x^2-4x+5}}]$$$$[\mathrm{use}x=\frac{t^2+4t-1}{2t}\wedge t>0]$$$$=\int \frac{{(t^2+1)}^2}{4t^3}dt=\int (\frac{t}{4}+\frac{1}{2t}+\frac{1}{4t^3})dt=$$$$=\frac{t^4-1}{8t^2}+\frac{1}{2}\ln t=$$$$=\frac{x-2}{2}\sqrt{x^2-4x+5}+\frac{1}{2}\ln (x-2+\sqrt{x^2-4x+5})+C$$

Answered by Dwaipayan Shikari last updated on 24/Nov/20

$$\int \sqrt{{(x-2)}^2+1}dx(x-2)=it\Rightarrow 1=i\frac{dt}{dx}$$$$=i\int \sqrt{1-t^2}dt=i\int cos\theta \sqrt{1-{sin}^2\theta }d\theta$$$$=i(\frac{\theta }{2}+\frac{sin2\theta }{4})=i(\frac{{sin}^{-1}t}{2}+\frac{t\sqrt{1-t^2}}{2})=i(\frac{{sin}^{-1}\left(\frac{x-2}{i}\right)}{2}+\frac{\frac{(x-2)}{i}\sqrt{1+{(x-2)}^2}}{2})$$$$=i\frac{{sin}^{-1}\left(\frac{x-2}{i}\right)}{2}+\frac{(x-2)\sqrt{x^2-4x+5}}{2}$$$$=\frac{1}{2}log(x-2+\sqrt{x^2-4x+5})+\frac{(x-2)}{2}\sqrt{x^2-4x+5}$$

Commented by Dwaipayan Shikari last updated on 24/Nov/20

$${isin}^{-1}\frac{(x-2)}{i}=t$$$$\frac{(x-2)}{i}=sin\left(\frac{t}{i}\right)\Rightarrow \left(\frac{x-2}{i}\right)=\frac{e^t-e^{-t}}{2i}\Rightarrow (2x-4)=e^t-e^{-t}$$$$e^t-e^{-t}=(2x-4)$$$$a-\frac{1}{a}=(2x-4)\Rightarrow a^2-(2x-4)a-1=0\Rightarrow a=\frac{(2x-4)+\sqrt{{(2x-4)}^2+4}}{2}$$$$a=x-2+\sqrt{x^2-4x+5}=e^t$$$$t=log(x-2+\sqrt{x^2-4x+5})$$

Answered by mathmax by abdo last updated on 24/Nov/20

$$\mathrm{A}=\int \sqrt{{\mathrm{x}}^2-4\mathrm{x}+5}\mathrm{dx}$$$${\mathrm{x}}^2-4\mathrm{x}+5={\mathrm{x}}^2-4\mathrm{x}+4+1={(\mathrm{x}-2)}^2+1\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}$$$$\mathrm{x}-2=\mathrm{sht}\Rightarrow \mathrm{A}=\int \sqrt{{(\mathrm{x}-2)}^2+1}\mathrm{dx}=\int \sqrt{1+{\mathrm{sh}}^2}\mathrm{t}\mathrm{ch}\left(\mathrm{t}\right)\mathrm{dt}$$$$=\int {\mathrm{ch}}^2\mathrm{t}\mathrm{dt}=\int \frac{\mathrm{ch}\left(2\mathrm{t}\right)+1}{2}\mathrm{dt}=\frac{\mathrm{t}}{2}+\frac{1}{4}\mathrm{sh}\left(2\mathrm{t}\right)+\mathrm{C}$$$$=\frac{\mathrm{t}}{2}+\frac{1}{2}\mathrm{sh}\left(\mathrm{t}\right)\mathrm{ch}\left(\mathrm{t}\right)+\mathrm{C}=\frac{1}{2}\mathrm{argsh}(\mathrm{x}-2)+\frac{\mathrm{x}-2}{2}\sqrt{1+{(\mathrm{x}-2)}^2}+\mathrm{C}$$$$=\frac{1}{2}\ln (\mathrm{x}-2+\sqrt{1+{(\mathrm{x}-2)}^2})+\frac{\mathrm{x}-2}{2}\sqrt{1+{(\mathrm{x}-2)}^2}+\mathrm{C}$$

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