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Question Number 132799 by Ñï= last updated on 16/Feb/21

$$\int \frac{dx}{1+x^4}$$$$=\frac{1}{2}\int \frac{dx}{1+{ix}^2}+\frac{1}{2}\int \frac{dx}{1-{ix}^2}$$$$=\frac{1}{2}{e}^{-i\frac{\pi }{4}}\int \frac{d\left(e^{i\frac{\pi }{4}}x\right)}{1+{\left(e^{i\frac{\pi }{4}}x\right)}^2}+\frac{1}{2}{e}^{i\frac{\pi }{4}}\int \frac{d\left(e^{-i\frac{\pi }{4}}\right)x}{1+{\left(e^{-i\frac{\pi }{4}}x\right)}^2}$$$$=\frac{1}{2}{e}^{-i\frac{\pi }{4}}{tan}^{-1}\left(e^{i\frac{\pi }{4}}x\right)+\frac{1}{2}{e}^{i\frac{\pi }{4}}{tan}^{-1}\left(e^{-i\frac{\pi }{4}}x\right)+C$$$$=Questionishowtocontinuefinishit?$$

Commented by Ar Brandon last updated on 16/Feb/21

$${\mathrm{x}}^4+1={({\mathrm{x}}^2+1)}^2-{2\mathrm{x}}^2={({\mathrm{x}}^2+1)}^2-{\left(\sqrt{2}\mathrm{x}\right)}^2$$$$=({\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1)({\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1)$$$$\frac{1}{{\mathrm{x}}^4+1}=\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1}+\frac{\mathrm{cx}+\mathrm{d}}{{\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1}$$$$\mathrm{a}+\mathrm{c}=0,\sqrt{2}\mathrm{a}+\mathrm{b}-\sqrt{2}\mathrm{c}+\mathrm{d}=0,\mathrm{a}+\sqrt{2}\mathrm{b}+\mathrm{c}-\sqrt{2}\mathrm{d}=0,\mathrm{b}+\mathrm{d}=1$$$$\mathrm{b}=\frac{1}{2}=\mathrm{d},\mathrm{a}=-\frac{1}{2\sqrt{2}},\mathrm{c}=\frac{1}{2\sqrt{2}}$$

Commented by MJS_new last updated on 16/Feb/21

$$\arctan (a+b\mathrm{i})=$$$$=\frac{1}{2}(\pi \mathrm{sign}a+\arctan \frac{b-1}{a}-\arctan \frac{b+1}{a})+\frac{\mathrm{i}}{4}\ln \frac{a^2+{(b+1)}^2}{a^2+{(b-1)}^2}$$$${\mathrm{that}}^{\prime }\mathrm{s}\mathrm{why}{\mathrm{I}}^{\prime }\mathrm{d}\mathrm{rather}\mathrm{not}...$$

Answered by MJS_new last updated on 16/Feb/21

$$\int \frac{dx}{x^4+1}=\frac{1}{4}\int \frac{\sqrt{2}x+2}{x^2+\sqrt{2}x+1}-\frac{\sqrt{2}x-2}{x^2-\sqrt{2}x+1}dx=$$$$=\int \frac{2x+\sqrt{2}}{4\sqrt{2}(x^2+\sqrt{2}x+1)}+\frac{1}{4(x^2+\sqrt{2}x+1)}-\frac{2x-\sqrt{2}}{4\sqrt{2}(x^2+\sqrt{2}x+1)}+\frac{1}{4(x^2-\sqrt{2}x+1)}dx=$$$$=\frac{\sqrt{2}}{8}\ln (x^2+\sqrt{2}x+1)+\frac{\sqrt{2}}{4}\arctan (\sqrt{2}x+1)-\frac{\sqrt{2}}{8}\ln (x^2-\sqrt{2}x+1)+\frac{\sqrt{2}}{4}\arctan (\sqrt{2}x-1)+C$$

Commented by Ñï= last updated on 16/Feb/21

$$Iwanttoknowhowtouse\mathrm{arc}tanfuctiontogetitback.$$$$ThismethodIalreadygetit.sir.$$

Answered by Ar Brandon last updated on 16/Feb/21

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

Answered by mathmax by abdo last updated on 16/Feb/21

$$\mathrm{for}\mathrm{that}\mathrm{find}\arctan (\mathrm{x}+\mathrm{iy})...$$

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