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Question Number 124000 by joki last updated on 30/Nov/20

Answered by liberty last updated on 30/Nov/20

$$\int \frac{dx}{(x-2)\sqrt{{(x-2)}^2+2^2}};[x-2=2\tan r]$$$$\mu \left(x\right)=\int \frac{2\sec {}^{2}rdr}{2\tan r\sqrt{4\sec {}^{2}r}}=\frac{1}{2}\int \frac{dr}{\sin r}$$$$\mu \left(x\right)=\frac{1}{2}\int \mathrm{cosec}rdr=\frac{1}{2}\ell n\mid \mathrm{cosec}r-\cot r\mid +c$$$$\mu \left(x\right)=\ell n\sqrt{\frac{\sqrt{x^2-4x+8}}{2}-\frac{2}{x-2}}+c$$

Answered by mathmax by abdo last updated on 30/Nov/20

$$\mathrm{A}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)\sqrt{{\mathrm{x}}^2-4\mathrm{x}+8}}\Rightarrow \mathrm{A}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)\sqrt{{(\mathrm{x}-2)}^2+4}}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}-2=2\mathrm{sht}\Rightarrow$$$$\mathrm{A}=\int \frac{2\mathrm{cht}}{2\mathrm{sht}2\mathrm{cht}}\mathrm{dt}=\int \frac{\mathrm{dt}}{2\mathrm{sh}\left(\mathrm{t}\right)}=\int \frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}$$$${=}_{{\mathrm{e}}^{\mathrm{t}}=\mathrm{z}}\int \frac{\mathrm{dz}}{\mathrm{z}(\mathrm{z}-{\mathrm{z}}^{-1})}=\int \frac{\mathrm{dz}}{{\mathrm{z}}^2-1}=\frac{1}{2}\int (\frac{1}{\mathrm{z}-1}-\frac{1}{\mathrm{z}+1})\mathrm{dz}$$$$=\frac{1}{2}\ln \mid \frac{\mathrm{z}-1}{\mathrm{z}+1}\mid +\mathrm{k}\mathrm{we}\mathrm{have}\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{x}-2}{2}\right)=\ln (\frac{\mathrm{x}-2}{2}+\sqrt{1+{\left(\frac{\mathrm{x}-2}{2}\right)}^2})$$$$\Rightarrow \mathrm{z}={\mathrm{e}}^{\mathrm{t}}=\frac{\mathrm{x}-2}{2}+\sqrt{1+{\left(\frac{\mathrm{x}-2}{2}\right)}^2}\Rightarrow$$$$\mathrm{A}=\frac{1}{2}\ln \mid \frac{\frac{\mathrm{x}-2}{2}+\frac{\sqrt{4+{(\mathrm{x}-2)}^2}}{2}-1\mid }{\frac{\mathrm{x}-2}{2}+\frac{\sqrt{4+{(\mathrm{x}-2)}^2}}{2}+1}\mid +\mathrm{k}$$$$=\frac{1}{2}\ln \mid \frac{\mathrm{x}-4+\sqrt{{(\mathrm{x}-2)}^2+4}}{\mathrm{x}+\sqrt{{(\mathrm{x}-2)}^2+4}}\mid +\mathrm{k}$$

Answered by Ar Brandon last updated on 30/Nov/20

$$\mathrm{t}=\frac{1}{\mathrm{x}-2}\Rightarrow \mathrm{dt}=-\frac{\mathrm{dx}}{{(\mathrm{x}-2)}^2}=-{\mathrm{t}}^2\mathrm{dx}$$$$\Rightarrow \mathcal{I}=-\int \frac{\mathrm{tdt}}{{\mathrm{t}}^2\sqrt{\frac{1}{{\mathrm{t}}^2}+4}}=-\int \frac{\mathrm{dt}}{\sqrt{1+{4\mathrm{t}}^2}}$$$$=-\frac{1}{2}\mathrm{Arcsinh}\left(2\mathrm{t}\right)+\mathcal{C}=-\frac{1}{2}\ln \mid 2\mathrm{t}+\sqrt{1+{4\mathrm{t}}^2}\mid +\mathcal{C}$$$$=-\frac{1}{2}\ln \mid \frac{2+\sqrt{{\mathrm{x}}^2-4\mathrm{x}+8}}{\mathrm{x}-2}\mid +\mathcal{C}$$

Commented by liberty last updated on 30/Nov/20

$$if2t=\cos u\Rightarrow I=-\int \frac{\frac{1}{2}(-\sin u)du}{\sqrt{1+\cos {}^{2}u}}$$$$howyougot\frac{1}{2}\mathrm{arc}\cos \left(2t\right)?$$

Commented by Ar Brandon last updated on 30/Nov/20

Found my mistake, thanks

Answered by MJS_new last updated on 30/Nov/20

$$\int \frac{dx}{(x-2)\sqrt{x^2-4x+8}}=$$$$[t=\frac{2+\sqrt{x^2-4x+8}}{x-2}\to dx=(-\frac{x^2}{2}+2x-4+\sqrt{x^2-4x+8})dt]$$$$=-\frac{1}{2}\int \frac{dt}{t}=-\frac{1}{2}\ln t=$$$$=-\frac{1}{2}\ln \mid \frac{2+\sqrt{x^2-4x+8}}{x-2}\mid +C$$

Answered by Dwaipayan Shikari last updated on 30/Nov/20

$$\sqrt{x^2-4x+8}=t\Rightarrow \frac{x-2}{\sqrt{x^2-4x+8}}=\frac{dt}{dx}$$$$=\int \frac{dt}{{(x-2)}^2}{x}^2-4x+8=t^2\Rightarrow {(x-2)}^2=(t^2-4)$$$$=\int \frac{1}{t^2-4}dt=\frac{1}{4}log\left(\frac{\sqrt{x^2-4x+8}-2}{\sqrt{x^2-4x+8}+2}\right)+C$$

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