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Question Number 120092 by danielasebhofoh last updated on 29/Oct/20

Commented by Lordose last updated on 29/Oct/20

SMALL EINSTEIN..?��

Answered by Lordose last updated on 29/Oct/20

$$\mathrm{\Omega }=\int \frac{{2\mathrm{x}}^2+5}{{\mathrm{x}}^3\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}\mathrm{dx}$$$$\mathrm{set}\mathrm{u}={\mathrm{x}}^2\Rightarrow \mathrm{du}=2\mathrm{xdx}$$$$\mathrm{\Omega }=\frac{1}{2}\int \frac{2\mathrm{u}+5}{{\mathrm{u}}^2\sqrt{{\mathrm{u}}^2+\mathrm{u}+1}}\mathrm{du}$$$$\mathrm{\Omega }=\frac{1}{2}\int \frac{\frac{1}{{\mathrm{u}}^2}(2\mathrm{u}+5)}{\sqrt{{(\mathrm{u}+\frac{1}{2})}^2+\frac{3}{4}}}\mathrm{du}$$$$\mathrm{set}\mathrm{u}+\frac{1}{2}=\mathrm{y}\Rightarrow \mathrm{du}=\mathrm{dy}$$$$\mathrm{\Omega }=\frac{1}{2}\int \frac{\frac{2}{(\mathrm{y}-\frac{1}{2})}+\frac{5}{{(\mathrm{y}-\frac{1}{2})}^2}}{\sqrt{{\mathrm{y}}^2+\frac{3}{4}}}\mathrm{dy}$$$$\mathrm{set}\mathrm{y}=\frac{\sqrt{3}}{2}\tan \theta \Rightarrow \mathrm{dy}=\frac{\sqrt{3}}{2}{\sec }^2\theta \mathrm{d}\theta$$$$\mathrm{\Omega }=\frac{1}{2}\int \frac{\frac{\sqrt{3}}{2}\sec \theta (\frac{2}{\frac{1}{2}(\sqrt{3}\tan \theta -1)}+\frac{5}{\frac{1}{4}{(\sqrt{3}\tan \theta -1)}^2})}{\frac{\sqrt{3}}{2}}\mathrm{d}\theta$$$$\mathrm{\Omega }=\frac{1}{2}\int (\frac{4\sec \theta }{\sqrt{3}\tan \theta -1}+\frac{20\sec \theta }{{(\sqrt{3}\tan \theta -1)}^2})\mathrm{d}\theta$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{from}\mathrm{here}$$

Answered by mathmax by abdo last updated on 29/Oct/20

$$\mathrm{A}=\int \frac{{2\mathrm{x}}^2+5}{{\mathrm{x}}^3\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}\mathrm{dx}\mathrm{changement}{\mathrm{x}}^2=\mathrm{t}\mathrm{give}$$$$\mathrm{A}=\int \frac{2\mathrm{t}+5}{\mathrm{t}\sqrt{\mathrm{t}}\sqrt{{\mathrm{t}}^2+\mathrm{t}+1}}\frac{\mathrm{dt}}{2\sqrt{\mathrm{t}}}=\int \frac{2\mathrm{t}+5}{{2\mathrm{t}}^2\sqrt{{\mathrm{t}}^2+\mathrm{t}+1}}\mathrm{dt}$$$$=\frac{1}{2}\int \frac{2\mathrm{t}+5}{{\mathrm{t}}^2\sqrt{{(\mathrm{t}+\frac{1}{2})}^2+\frac{3}{4}}}{=}_{\mathrm{t}+\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{shu}}\frac{1}{2}\int \frac{2(\frac{\sqrt{3}}{2}\mathrm{shu}-\frac{1}{2})+5}{{(\frac{\sqrt{3}}{2}\mathrm{shu}-\frac{1}{2})}^2\sqrt{\frac{3}{4}}\mathrm{chu}}\frac{\sqrt{3}}{2}\mathrm{chudu}$$$$=2\int \frac{\sqrt{3}\mathrm{shu}+4}{{(\sqrt{3}\mathrm{shu}-1)}^2}\mathrm{du}=2\int \frac{\sqrt{3}\frac{{\mathrm{e}}^{\mathrm{u}}-{\mathrm{e}}^{-\mathrm{u}}}{2}+4}{{(\sqrt{3}\frac{{\mathrm{e}}^{\mathrm{u}}-{\mathrm{e}}^{-\mathrm{u}}}{2}-1)}^2}\mathrm{du}$$$$=4\int \frac{\sqrt{3}{\mathrm{e}}^{\mathrm{u}}-\sqrt{3}{\mathrm{e}}^{-\mathrm{u}}+8}{{(\sqrt{3}({\mathrm{e}}^{\mathrm{u}}-{\mathrm{e}}^{-\mathrm{u}})-2)}^2}\mathrm{du}=4\int \frac{\sqrt{3}{\mathrm{e}}^{\mathrm{u}}-\sqrt{3}{\mathrm{e}}^{-\mathrm{u}}+8}{3{({\mathrm{e}}^{\mathrm{u}}-{\mathrm{e}}^{-\mathrm{u}})}^2-4\sqrt{3}({\mathrm{e}}^{\mathrm{u}}-{\mathrm{e}}^{-\mathrm{u}})+4}\mathrm{du}$$$$=4\int \frac{\sqrt{3}{\mathrm{e}}^{\mathrm{u}}-\sqrt{3}{\mathrm{e}}^{-\mathrm{u}}+8}{3({\mathrm{e}}^{2\mathrm{u}}-2+{\mathrm{e}}^{-2\mathrm{u}})-4\sqrt{3}{\mathrm{e}}^{\mathrm{u}}+4\sqrt{3}{\mathrm{e}}^{-\mathrm{u}}+4}\mathrm{du}$$$$=4\int \frac{\sqrt{3}{\mathrm{e}}^{\mathrm{u}}-\sqrt{3}{\mathrm{e}}^{-\mathrm{u}}+8}{{3\mathrm{e}}^{2\mathrm{u}}-2+{3\mathrm{e}}^{-2\mathrm{u}}-4\sqrt{3}{\mathrm{e}}^{\mathrm{u}}+4\sqrt{3}{\mathrm{e}}^{-\mathrm{u}}}\mathrm{du}$$$$=4\int \frac{\sqrt{3}{\mathrm{e}}^{3\mathrm{u}}-\sqrt{3}{\mathrm{e}}^{\mathrm{u}}+{8\mathrm{e}}^{2\mathrm{u}}}{{3\mathrm{e}}^{4\mathrm{u}}+3-4\sqrt{3}{\mathrm{e}}^{3\mathrm{u}}+4\sqrt{3}\mathrm{u}-{2\mathrm{e}}^{2\mathrm{u}}}\mathrm{du}$$$${=}_{{\mathrm{e}}^{\mathrm{u}}=\mathrm{z}}4\int \frac{\sqrt{3}{\mathrm{z}}^3+{8\mathrm{z}}^2-\sqrt{3}\mathrm{z}}{{3\mathrm{z}}^4-4\sqrt{3}{\mathrm{z}}^3-{2\mathrm{z}}^2+4\sqrt{3}\mathrm{z}+3}\frac{\mathrm{dz}}{\mathrm{z}}$$$$=4\int \frac{\sqrt{3}{\mathrm{z}}^2+8\mathrm{z}-\sqrt{3}}{{3\mathrm{z}}^4-4\sqrt{3}{\mathrm{z}}^3-{2\mathrm{z}}^2+4\sqrt{3}\mathrm{z}+3}\mathrm{dx}\mathrm{rest}\mathrm{decoposition}...\mathrm{be}\mathrm{continued}$$$$.......$$

Answered by TANMAY PANACEA last updated on 29/Oct/20

$$\int \frac{\frac{2}{x^3}+\frac{5}{x^5}}{\sqrt{1+\frac{1}{x^2}+\frac{1}{x^4}}}dx$$$$t^2=1+x^{-2}+x^{-4}$$$$2tdt=(\frac{-2}{x^3}+\frac{-4}{x^5})dx$$$$-2tdt=(\frac{2}{x^3}+\frac{4}{x^5})dx$$$$\int \frac{\frac{2}{x^3}+\frac{4}{x^5}}{\sqrt{1+\frac{1}{x^2}+\frac{1}{x^4}}}dx+\int \frac{\frac{1}{x^5}}{\sqrt{1+\frac{1}{x^2}+\frac{1}{x^4}}}dx$$$$\mathit{I}={\mathit{I}}_1+I_2$$$$I_1=\int \frac{-2tdt}{t}=-2t=-2\left(\sqrt{1+\frac{1}{x^2}+\frac{1}{x^4}}\right)+c_1$$$$I_2=\int \frac{dx}{x^5\sqrt{\frac{x^4+x^2+1}{x^4}}}$$$$=\int \frac{dx}{x^3\sqrt{x^4+x^2+1}}\to k=x^{2}dk=2xdx$$$$\frac{1}{2}\int \frac{2xdx}{x^4\sqrt{x^4+x^2+1}}$$$$\frac{1}{2}\int \frac{dk}{k^2\sqrt{k^2+k+1}}$$$$Wait$$$$p=\frac{1}{k}\to dp=\frac{-dk}{k^2}$$$$\frac{1}{2}\int \frac{-dp}{\sqrt{\frac{1}{p^2}+\frac{1}{p}+1}}$$$$=\left(\frac{-1}{2}\right)\int \frac{pdp}{\sqrt{p^2+p+1}}$$$$=\left(\frac{-1}{4}\right)\int \frac{2p+1-1}{\sqrt{p^2+p+1}}dp$$$$=\left(\frac{-1}{4}\right)\int \frac{d(p^2+p+1)}{\sqrt{p^2+p+1}}+\frac{1}{4}\int \frac{dp}{\sqrt{p^2+2p\times \frac{1}{2}+\frac{1}{4}+\frac{3}{4}}}$$$$=\left(\frac{-1}{4}\right)\times \frac{\sqrt{p^2+p+1}}{\frac{1}{2}}+\int \frac{dp}{\sqrt{{(p+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$$$$=\left(\frac{-1}{2}\right)\sqrt{p^2+p+1}+ln\{(p+\frac{1}{2})+\sqrt{{(p+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+c_2$$$$\mathit{p}=\frac{1}{\mathit{k}}\mathit{a}\mathit{n}\mathit{d}\mathit{k}={\mathit{x}}^2\mathit{s}\mathit{o}\mathit{r}\mathit{e}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}\mathit{p}\mathit{b}\mathit{y}=\frac{1}{{\mathit{x}}^2}$$$${\mathit{I}}_2=\left(\frac{-1}{2}\right)\sqrt{\frac{1}{x^4}+\frac{1}{x^2}+1}+ln\{(\frac{1}{x^2}+\frac{1}{2})+\sqrt{{(\frac{1}{x^2}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+c_2$$$$nowpleaseadd{I}_{1}and{I}_2\to I=I_1+I_2$$

Commented by peter frank last updated on 30/Oct/20

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

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