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Question Number 109246 by peter frank last updated on 22/Aug/20

Answered by Ar Brandon last updated on 22/Aug/20

$$\mathrm{a}.\mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+4\mathrm{x}+2}}=\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{(\mathrm{x}+2)}^2-2}}$$$$\mathrm{x}+2=\sqrt{2}\mathrm{ch}\theta \Rightarrow \mathrm{dx}=\sqrt{2}\mathrm{sh}\theta \mathrm{d}\theta$$$$\mathrm{I}=\int \frac{\mathrm{sh}\theta \mathrm{d}\theta }{(\sqrt{2}\mathrm{ch}\theta -1)\sqrt{{\mathrm{sh}}^2\theta }}=\pm \int \frac{\mathrm{d}\theta }{\sqrt{2}\mathrm{ch}\theta -1}$$$$\mathrm{ch}\theta =\frac{1+{\mathrm{t}}^2}{1-{\mathrm{t}}^2},\mathrm{t}=\tanh \left(\frac{\theta }{2}\right)$$

Answered by Ar Brandon last updated on 22/Aug/20

$$\mathrm{b}.\mathrm{J}=\int \frac{\mathrm{dx}}{({\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+2}},\mathrm{x}=\tan \theta$$$$\mathrm{J}=\int \frac{\mathrm{d}\theta }{\sqrt{{\tan }^2\theta +2}}=\int \frac{\cos \theta \mathrm{d}\theta }{\sqrt{{\sin }^2\theta +{2\cos }^2\theta }}$$$$=\int \frac{\cos \theta }{\sqrt{2-{\sin }^2\theta }}\mathrm{d}\theta =\mathrm{Arcsin}\left(\frac{\sin \theta }{\sqrt{2}}\right)+\mathrm{C}$$

Answered by Ar Brandon last updated on 22/Aug/20

$$\mathrm{a}.$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+4\mathrm{x}+2}},\mathrm{u}=\frac{1}{\mathrm{x}+1}\Rightarrow \mathrm{du}=\frac{-\mathrm{dx}}{{(\mathrm{x}+1)}^2}$$$$\mathrm{I}=\int \frac{-{(\mathrm{x}+1)}^2\mathrm{u}}{\sqrt{{(\mathrm{x}+2)}^2-2}}\mathrm{du}=-\int \frac{\mathrm{udu}}{\sqrt{{(\mathrm{u}+{\mathrm{u}}^2)}^2-{2\mathrm{u}}^4}}$$$$=-\int \frac{\mathrm{du}}{\sqrt{{(1+\mathrm{u})}^2-{2\mathrm{u}}^2}}=-\int \frac{\mathrm{du}}{\sqrt{1+2\mathrm{u}-{\mathrm{u}}^2}}$$$$=-\int \frac{\mathrm{du}}{\sqrt{2-{(\mathrm{u}-1)}^2}}=\mathrm{Arcos}\left(\frac{\mathrm{u}-1}{\sqrt{2}}\right)+\mathrm{C}$$

Commented by peter frank last updated on 22/Aug/20

$$\mathrm{thank}\mathrm{you}$$

Answered by mathmax by abdo last updated on 22/Aug/20

$$\mathrm{a})\mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+4\mathrm{x}+2}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{(\mathrm{x}+2)}^2-2}}$$$${=}_{\mathrm{x}+2=\sqrt{2}\mathrm{ch}\left(\mathrm{t}\right)}\int \frac{\sqrt{2}\mathrm{sh}\left(\mathrm{t}\right)}{(-2+\sqrt{2}\mathrm{cht}+1)\sqrt{2}\mathrm{sh}\left(\mathrm{t}\right)}\mathrm{dt}$$$$=\int \frac{\mathrm{dt}}{-1+\sqrt{2}\frac{{\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}}{2}}=\int \frac{2\mathrm{dt}}{-2+\sqrt{2}{\mathrm{e}}^{\mathrm{t}}+\sqrt{2}{\mathrm{e}}^{-\mathrm{t}}}$$$${=}_{{\mathrm{e}}^{\mathrm{t}}=\mathrm{u}}2\int \frac{\mathrm{du}}{\mathrm{u}(-2+\sqrt{2}\mathrm{u}+\sqrt{2}{\mathrm{u}}^{-1})}=2\int \frac{\mathrm{du}}{-2\mathrm{u}+\sqrt{2}{\mathrm{u}}^2+\sqrt{2}}$$$$=\sqrt{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2-\sqrt{2}\mathrm{u}+1}=\sqrt{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2-2\frac{1}{\sqrt{2}}\mathrm{u}+\frac{1}{2}+\frac{1}{2}}$$$$=\sqrt{2}\int \frac{\mathrm{du}}{{(\mathrm{u}-\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}{=}_{\mathrm{u}-\frac{1}{\sqrt{2}}=\frac{\mathrm{z}}{\sqrt{2}}}\sqrt{2}\int \frac{\mathrm{dz}}{\sqrt{2}.\frac{1}{2}(1+{\mathrm{z}}^2)}=\int \frac{2\mathrm{dz}}{1+{\mathrm{z}}^2}$$$$=2\mathrm{arctanz}+\mathrm{C}=2\arctan (\mathrm{u}\sqrt{2}-1)+\mathrm{C}$$$$\mathrm{we}\mathrm{have}\mathrm{t}=\mathrm{argch}\left(\frac{\mathrm{x}+2}{\sqrt{2}}\right)=\ln (\frac{\mathrm{x}+2}{\sqrt{2}}+\sqrt{{\left(\frac{\mathrm{x}+2}{\sqrt{2}}\right)}^2-1})\Rightarrow$$$$\mathrm{u}={\mathrm{e}}^{\mathrm{t}}=\frac{\mathrm{x}+2}{\sqrt{2}}+\sqrt{{\left(\frac{\mathrm{x}+2}{\sqrt{2}}\right)}^2-1}\Rightarrow$$$$\mathrm{I}=2\arctan \left(\sqrt{2}\{\frac{\mathrm{x}+2}{\sqrt{2}}+\sqrt{{\left(\frac{\mathrm{x}+2}{\sqrt{2}}\right)}^2-1}\}\right)+\mathrm{C}$$

Commented by mathmax by abdo last updated on 22/Aug/20

$$\mathrm{I}=2\arctan (\mathrm{x}+1+\sqrt{{(\mathrm{x}+2)}^2-2})+\mathrm{C}$$

Commented by peter frank last updated on 22/Aug/20

$$\mathrm{thank}\mathrm{you}$$

Commented by mathmax by abdo last updated on 22/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by mathmax by abdo last updated on 22/Aug/20

$$\mathrm{b})\mathrm{A}=\int \frac{\mathrm{dx}}{({\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+2}}\Rightarrow \mathrm{A}{=}_{\mathrm{x}=\sqrt{2}\mathrm{sht}}\int \frac{\sqrt{2}\mathrm{ch}\left(\mathrm{t}\right)}{({2\mathrm{sh}}^2\mathrm{t}+1)\sqrt{2}\mathrm{ch}\left(\mathrm{t}\right)}\mathrm{dt}$$$$=\int \frac{\mathrm{dt}}{2.\frac{\mathrm{ch}\left(2\mathrm{t}\right)-1}{2}+1}=\int \frac{\mathrm{dt}}{\mathrm{ch}\left(2\mathrm{t}\right)}{=}_{2\mathrm{t}=\mathrm{u}}\int \frac{\mathrm{du}}{2\mathrm{ch}\left(\mathrm{u}\right)}$$$$=\int \frac{\mathrm{du}}{{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}}}{=}_{{\mathrm{e}}^{\mathrm{u}}=\mathrm{z}}\int \frac{\mathrm{dz}}{\mathrm{z}(\mathrm{z}+{\mathrm{z}}^{-1})}=\int \frac{\mathrm{dz}}{{\mathrm{z}}^2+1}=\mathrm{arctanz}+\mathrm{c}$$$$=\arctan \left(\mathrm{z}\right)+\mathrm{c}=\arctan \left({\mathrm{e}}^{\mathrm{u}}\right)+\mathrm{c}=\arctan \left({\mathrm{e}}^{2\mathrm{t}}\right)+\mathrm{c}$$$$\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{x}}{\sqrt{2}}\right)=\ln (\frac{\mathrm{x}}{\sqrt{2}}+\sqrt{1+\frac{{\mathrm{x}}^2}{2}})\Rightarrow {\mathrm{e}}^{2\mathrm{t}}={(\frac{\mathrm{x}}{2}+\sqrt{1+\frac{{\mathrm{x}}^2}{2}})}^2\Rightarrow$$$$\mathrm{A}=\arctan \left\{{(\frac{\mathrm{x}}{2}+\sqrt{1+\frac{{\mathrm{x}}^2}{2}})}^2\right\}+\mathrm{C}$$$$$$

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