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Question Number 126510 by benjo_mathlover last updated on 21/Dec/20

$$\int \frac{dx}{(x-2)\sqrt{x^2+4x+8}}?$$

Answered by liberty last updated on 21/Dec/20

Answered by Ar Brandon last updated on 21/Dec/20

$$\mathcal{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)\sqrt{{\mathrm{x}}^2+4\mathrm{x}+8}}$$$$\mathrm{Let}\frac{1}{\mathrm{x}-2}=\mathrm{u}\Rightarrow -\frac{\mathrm{dx}}{{(\mathrm{x}-2)}^2}=\mathrm{du}\Rightarrow \mathrm{dx}=-\frac{\mathrm{du}}{{\mathrm{u}}^2}$$$$\mathrm{x}=\frac{1}{\mathrm{u}}+2,{\mathrm{x}}^2=\frac{1}{{\mathrm{u}}^2}+\frac{4}{\mathrm{u}}+4$$$$\mathcal{I}=-\int \frac{\mathrm{u}}{\sqrt{\frac{1}{{\mathrm{u}}^2}+\frac{8}{\mathrm{u}}+20}}\cdot \frac{\mathrm{du}}{{\mathrm{u}}^2}$$$$=-\int \frac{\mathrm{du}}{\sqrt{1+8\mathrm{u}+{20\mathrm{u}}^2}}=\mp \frac{1}{\sqrt{20}}\int \frac{\mathrm{du}}{\sqrt{{\mathrm{u}}^2+\frac{2\mathrm{u}}{5}+\frac{1}{20}}}$$$$=\mp \frac{1}{\sqrt{20}}\int \frac{\mathrm{du}}{\sqrt{{(\mathrm{u}+\frac{1}{5})}^2+\frac{1}{100}}}$$$$=\mp \frac{1}{\sqrt{20}}\ln \mid (\mathrm{u}+\frac{1}{5})+\sqrt{{\mathrm{u}}^2+\frac{2\mathrm{u}}{5}+\frac{1}{20}}\mid +C$$$$=\mp \frac{1}{\sqrt{20}}\ln \mid (\frac{1}{\mathrm{x}-2}+\frac{1}{5})+\sqrt{\frac{{\mathrm{x}}^2+4\mathrm{x}+8}{20}}\mid +C$$

Answered by mathmax by abdo last updated on 21/Dec/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)\sqrt{{\mathrm{x}}^2+4\mathrm{x}+8}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)\sqrt{{(\mathrm{x}+2)}^2+4}}$$$${=}_{\mathrm{x}+2=2\mathrm{sht}}\int \frac{2\mathrm{ch}\left(\mathrm{t}\right)}{(2\mathrm{sht}-4)2\mathrm{cht}}\mathrm{dt}=\frac{1}{2}\int \frac{\mathrm{dt}}{\mathrm{sht}-2}$$$$=\frac{1}{2}\int \frac{\mathrm{dt}}{\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2}-2}=\int \frac{\mathrm{dt}}{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}-4}{=}_{{\mathrm{e}}^{\mathrm{t}}=\mathrm{z}}\int \frac{\mathrm{dz}}{\mathrm{z}(\mathrm{z}-{\mathrm{z}}^{-1}-4)}$$$$=\int \frac{\mathrm{dz}}{{\mathrm{z}}^2-1-4\mathrm{z}}=\int \frac{\mathrm{dz}}{{\mathrm{z}}^2-4\mathrm{z}-1}$$$${\mathrm{\Delta }}^{{}^{\prime }}=4+1=5\Rightarrow {\mathrm{z}}_1=2+\sqrt{5}\mathrm{and}{\mathrm{z}}_2=2-\sqrt{5}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dz}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}$$$$=\frac{1}{2\sqrt{5}}\int (\frac{1}{\mathrm{z}-{\mathrm{z}}_1}-\frac{1}{\mathrm{z}-{\mathrm{z}}_2})\mathrm{dz}=\frac{1}{2\sqrt{5}}\ln \mid \frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{z}-{\mathrm{z}}_2}\mid +\mathrm{C}$$$$=\frac{1}{2\sqrt{5}}\ln \mid \frac{{\mathrm{e}}^{\mathrm{t}}-2-\sqrt{5}}{{\mathrm{e}}^{\mathrm{t}}-2+\sqrt{5}}\mid +\mathrm{C}\mathrm{we}\mathrm{have}\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{x}+2}{2}\right)$$$$=\ln (\frac{\mathrm{x}+2}{2}+\sqrt{1+{\left(\frac{\mathrm{x}+2}{2}\right)}^2})\Rightarrow {\mathrm{e}}^{\mathrm{t}}=\frac{\mathrm{x}}{2}+1+\sqrt{1+{(\frac{\mathrm{x}}{2}+1)}^2}$$$$\mathrm{I}=\frac{1}{2\sqrt{5}}\ln \mid \frac{\frac{\mathrm{x}}{2}+\sqrt{1+{(\frac{\mathrm{x}}{2}+1)}^2}-1-\sqrt{5}}{\frac{\mathrm{x}}{2}+\sqrt{1+{(\frac{\mathrm{x}}{2}+1)}^2}-1+\sqrt{5}}\mid +\mathrm{C}$$

Answered by MJS_new last updated on 21/Dec/20

$$\int \frac{dx}{(x-2)\sqrt{x^2+4x+8}}=$$$$[t=\frac{x+2+\sqrt{x^2+4x+8}}{2}\to dx=\frac{2\sqrt{x^2+4x+8}}{x+2+\sqrt{x^2+4x+8}}]$$$$=\int \frac{dt}{t^2-4t-1}=\frac{\sqrt{5}}{10}\int (\frac{1}{t-2-\sqrt{5}}-\frac{1}{t-2+\sqrt{5}})dt=$$$$=\frac{\sqrt{5}}{10}\ln \frac{t-2-\sqrt{5}}{t-2+\sqrt{5}}=\frac{\sqrt{5}}{10}\ln \mid \frac{2(2+\sqrt{5})(x+3)-(5+2\sqrt{5})\sqrt{x^2+4x+8}}{x-2}\mid +C$$

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