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Question Number 130287 by naka3546 last updated on 24/Jan/21

$$\int \frac{x}{x^4+1}dx=?$$

Answered by mathmax by abdo last updated on 24/Jan/21

$$\mathrm{I}=\int \frac{\mathrm{xdx}}{{\mathrm{x}}^4+1}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{x}}^4+1}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{({\mathrm{x}}^2+1)}^2-{2\mathrm{x}}^2}=\frac{\mathrm{x}}{({\mathrm{x}}^2+1-\sqrt{2}\mathrm{x})({\mathrm{x}}^2+1+\sqrt{2}\mathrm{x})}$$$$=\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1}+\frac{\mathrm{mx}+\mathrm{n}}{{\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1}\mathrm{we}\mathrm{see}[\mathrm{that}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{2\sqrt{2}}(\frac{1}{{\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1}-\frac{1}{{\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1})$$$$\Rightarrow \int \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{2\sqrt{2}}\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-2\frac{\sqrt{2}}{2}\mathrm{x}+\frac{1}{2}+\frac{1}{2}}-\frac{1}{2\sqrt{2}}\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+2\frac{\sqrt{2}}{2}\mathrm{x}+\frac{1}{2}+\frac{1}{2}}$$$$=\frac{1}{2\sqrt{2}}\int \frac{\mathrm{dx}}{{(\mathrm{x}-\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}(\to \mathrm{x}-\frac{1}{\sqrt{2}}=\frac{\mathrm{u}}{\sqrt{2}})-\frac{1}{2\sqrt{2}}\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}(\to \mathrm{x}+\frac{1}{\sqrt{2}}=\frac{\mathrm{v}}{\sqrt{2}})$$$$=\frac{1}{2\sqrt{2}}\int \frac{\mathrm{du}}{\sqrt{2}\left(\frac{1}{2}({\mathrm{u}}^2+1)\right)}-\frac{1}{2\sqrt{2}}\int \frac{\mathrm{dv}}{\sqrt{2}\frac{1}{2}({\mathrm{v}}^2+1)}$$$$=\frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+1}-\frac{1}{2}\int \frac{\mathrm{dv}}{{\mathrm{v}}^2+1}=\frac{1}{2}\arctan \left(\mathrm{u}\right)-\frac{1}{2}\mathrm{arctanv}+\mathrm{C}$$$$=\frac{1}{2}\arctan (\sqrt{2}\mathrm{x}-1)-\frac{1}{2}\arctan (\sqrt{2}\mathrm{x}+1)+\mathrm{C}$$$$$$

Commented by naka3546 last updated on 24/Jan/21

$$\mathrm{thanks},sir.$$$$$$$$Howabout\int \frac{x}{x^4+3}dx?$$

Commented by Ar Brandon last updated on 24/Jan/21

$$\int \frac{\mathrm{x}}{{\mathrm{x}}^4+3}\mathrm{dx}=\frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+3}=\frac{{\tan }^{-1}\left({\mathrm{x}}^2/\sqrt{3}\right)}{2\sqrt{3}}+\mathcal{C}$$

Commented by mathmax by abdo last updated on 24/Jan/21

$$\int \frac{\mathrm{xdx}}{{\mathrm{x}}^4+3}=\int \frac{\mathrm{xdx}}{3(\frac{{\mathrm{x}}^4}{3}+1)}=\int \frac{\mathrm{xdx}}{3({\left(\frac{\mathrm{x}}{\left({}^4\sqrt{3}\right)}\right)}^4+1)}$$$${=}_{\frac{\mathrm{x}}{\left({}^4\sqrt{3}\right)}=\mathrm{t}}\frac{1}{3}\int \frac{\left({}^4\sqrt{3}\right)\mathrm{t}}{1+{\mathrm{t}}^4}\left({}^4\sqrt{3}\right)\mathrm{dt}=\frac{3^{\frac{2}{4}}}{3}\int \frac{\mathrm{tdt}}{1+{\mathrm{t}}^4}=....$$

Commented by mathmax by abdo last updated on 26/Jan/21

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

Answered by Ar Brandon last updated on 24/Jan/21

$$\mathcal{I}=\int \frac{\mathrm{x}}{{\mathrm{x}}^4+1}\mathrm{dx},{\mathrm{x}}^2=\mathrm{u}$$$$=\frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2+1}=\frac{{\tan }^{-1}\left(\mathrm{u}\right)}{2}+\mathcal{C}$$$$=\frac{{\tan }^{-1}\left({\mathrm{x}}^2\right)}{2}+\mathcal{C}$$

Answered by mnjuly1970 last updated on 24/Jan/21

$$x^2=u$$$$xdx=\frac{du}{2}$$$$\varphi =\int \frac{du}{2(u^2+1)}=\frac{1}{2}{tan}^{-1}\left(x^2\right)+C...$$

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