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Question Number 122614 by john santu last updated on 18/Nov/20

Answered by liberty last updated on 18/Nov/20

$$F=\int \frac{dx}{\sqrt{x^2+1}(x^2+x+1)}$$$$Bytrigonometrysubstitutionmethod$$$$letx=\tan \theta$$$$F=\int \frac{\sec {}^2\theta d\theta }{\sqrt{\tan {}^2\theta +1}(\tan {}^2\theta +\tan \theta +1)}$$$$F=\int \frac{\sec \theta d\theta }{\sec {}^2\theta +\tan \theta }=\int \frac{\sec \theta \cos {}^2\theta d\theta }{1+\sin \theta \cos \theta }$$$$F=\int \frac{\cos \theta d\theta }{1+\sin \theta \cos \theta }=\int \frac{2\cos \theta }{2+\sin 2\theta }d\theta$$$$F=\int \frac{(\cos \theta +\sin \theta )}{3-{(\sin \theta -\cos \theta )}^2}+\int \frac{(\cos \theta -\sin \theta )}{1+{(\sin \theta +\cos \theta )}^2}d\theta$$$$nowletw=\cos \theta +\sin \theta \Rightarrow dw=(\cos \theta -\sin \theta )d\theta$$$$F=\int \frac{dw}{3-w^2}+\int \frac{dw}{1+w^2}$$$$F=\frac{\sqrt{3}}{6}\ell n\mid \frac{w+\sqrt{3}}{\sqrt{3}-w}\mid +\arctan \left(w\right)+c$$$$F=\frac{\sqrt{3}}{6}\ell n\mid \frac{\cos \theta -\sin \theta +\sqrt{3}}{\sqrt{3}-\cos \theta +\sin \theta }\mid +\arctan (\cos \theta +\sin \theta )+c$$$$F=\frac{\sqrt{3}}{6}\ell n\mid \frac{1-x+\sqrt{3(x^2+1)}}{x-1+\sqrt{3(x^2+1)}}\mid +\arctan \left(\frac{x+1}{\sqrt{x^2+1}}\right)+c.$$$$$$

Commented by john santu last updated on 18/Nov/20

$$thankyouallmaster$$

Commented by MJS_new last updated on 18/Nov/20

$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{lost}\mathrm{a}‘‘{-}^{″}\mathrm{somewhere}?$$

Commented by liberty last updated on 18/Nov/20

in what part sir?

Commented by MJS_new last updated on 18/Nov/20

$$\mathrm{not}\mathrm{sure}.\mathrm{but}\frac{d}{dx}\left[\mathrm{F}\right]=-\frac{1}{\sqrt{x^2+1}(x^2+x+1)}$$$$\mathrm{if}\mathrm{I}\mathrm{am}\mathrm{right}$$

Answered by Olaf last updated on 18/Nov/20

$$x=\sinh u$$$$\mathrm{I}=\int \frac{\cosh udu}{\sqrt{{\sinh }^{2}u+1}({\sinh }^{2}u+\sinh u+1)}$$$$\mathrm{I}=\int \frac{du}{{\sinh }^{2}u+\sinh u+1}$$$$\mathrm{I}=\int \frac{du}{\frac{e^{2u}-2+e^{-2u}}{4}+\frac{e^u-e^{-u}}{2}+1}$$$$\mathrm{I}=\int \frac{4du}{e^{2u}-2+e^{-2u}+2e^u-2e^{-u}+4}$$$$\mathrm{I}=\int \frac{4du}{e^{2u}+2e^u+2-2e^{-u}+e^{-2u}}$$$$\mathrm{I}=\int \frac{4e^{2u}du}{e^{4u}+2e^{3u}+2e^{2u}-2e^u+1}$$$$\mathrm{Let}\phi =e^u$$$$\mathrm{I}=\int \frac{4{\phi }^2\frac{d\phi }{\phi }}{{\phi }^4+2{\phi }^3+2{\phi }^2-2\phi +1}$$$$\mathrm{I}=\int \frac{4\phi d\phi }{{\phi }^4+2{\phi }^3+2{\phi }^2-2\phi +1}$$$$...4\mathrm{complex}\mathrm{roots}...$$

Answered by MJS_new last updated on 18/Nov/20

$$\int \frac{dx}{(x^2+x+1)\sqrt{x^2+1}}=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}dt]$$$$=4\int \frac{t}{t^4+2t^3+2t^2-2t+1}dt=$$$$=4\int \frac{t}{(t^2+(1-\sqrt{3})t+2-\sqrt{3})(t^2+(1+\sqrt{3})t+2+\sqrt{3})}dt=$$$$=\frac{\sqrt{3}}{12}\int (\frac{t+2-\sqrt{3}}{t^2+(1-\sqrt{3})t+2-\sqrt{3}}-\frac{t+2+\sqrt{3}}{t^2+(1+\sqrt{3})t+2+\sqrt{3}})dt=$$$$=\frac{\sqrt{3}}{24}\ln \frac{t^2+(1-\sqrt{3})t+2-\sqrt{3}}{t^2+(1+\sqrt{3})t+2+\sqrt{3}}+\frac{1}{4}(\arctan ((1+\sqrt{3})t-1)+\arctan ((1-\sqrt{3})t-1)$$$$...$$

Answered by mathmax by abdo last updated on 18/Nov/20

$$\mathrm{A}=\int \frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^2}({\mathrm{x}}^2+\mathrm{x}+1)}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{sht}\Rightarrow$$$$\mathrm{A}=\int \frac{\mathrm{cht}}{\mathrm{cht}({\mathrm{sh}}^2\mathrm{t}+\mathrm{sht}+1)}\mathrm{dt}=\int \frac{\mathrm{dt}}{\frac{\mathrm{ch}\left(2\mathrm{t}\right)-1}{2}+\mathrm{sh}\left(\mathrm{t}\right)+1}$$$$=\int \frac{2\mathrm{dt}}{\mathrm{ch}\left(2\mathrm{t}\right)-1+2\mathrm{sh}\left(\mathrm{t}\right)+2}=\int \frac{2\mathrm{dt}}{\frac{{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}}{2}+{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}+1}$$$$=\int \frac{4\mathrm{dt}}{{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}+{2\mathrm{e}}^{\mathrm{t}}-{2\mathrm{e}}^{-\mathrm{t}}+2}{=}_{{\mathrm{e}}^{\mathrm{t}}=\mathrm{u}}\int \frac{4\mathrm{du}}{\mathrm{u}({\mathrm{u}}^2+{\mathrm{u}}^{-2}+2\mathrm{u}-{2\mathrm{u}}^{-1}+2)}$$$$=\int \frac{4\mathrm{du}}{{\mathrm{u}}^3+{\mathrm{u}}^{-1}+{2\mathrm{u}}^2-2+2\mathrm{u}}=\int \frac{4\mathrm{u}\mathrm{du}}{{\mathrm{u}}^4+1+{2\mathrm{u}}^3-2\mathrm{u}+{2\mathrm{u}}^2}$$$$=\int \frac{4\mathrm{udu}}{{\mathrm{u}}^4+{2\mathrm{u}}^3+{2\mathrm{u}}^2-2\mathrm{u}+1}\mathrm{rest}\mathrm{decomposition}\mathrm{of}$$$$\mathrm{f}\left(\mathrm{u}\right)=\frac{4\mathrm{u}}{{\mathrm{u}}^4+{2\mathrm{u}}^3+{2\mathrm{u}}^2-2\mathrm{u}+1}...\mathrm{be}\mathrm{continued}....$$

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