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Question Number 193538 by cortano12 last updated on 16/Jun/23

$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^6+1}=?$$

Answered by Subhi last updated on 16/Jun/23

$$\int \frac{1}{(x^2+1)(x^4-x^2+1)}\Rrightarrow \int \frac{a}{x^2+1}+\frac{{bx}^2+cx+d}{x^4-x^2+1}$$$${ax}^4-{ax}^2+a+{bx}^4+{bx}^2+{cx}^3+cx+{dx}^2+d$$$$(a+b)x^4+(b-a+d)x^2+{cx}^3+cx+(d+a)$$$$c=0\Rrightarrow d+a=1\Rrightarrow a+b=0\Rrightarrow b-a+d=0$$$$2d+b=1\Rrightarrow d+2b=0$$$$b=\frac{-1}{3}\Rrightarrow d=\frac{2}{3}\Rrightarrow a=\frac{1}{3}$$$$\frac{1}{3}\int \frac{1}{x^2+1}+\frac{1}{3}\int \frac{-x^2+2}{x^4-x^2+1}\Rrightarrow \frac{{tan}^{-1}\left(x\right)}{3}-\frac{1}{3}\int \frac{x^2-2}{x^4-x^2+1}$$$$\int \frac{x^2-2}{x^4-x^2+1}\looparrowright \int \frac{x^2-2}{{(x^2+1)}^2-3x^2}\looparrowright \int \frac{x^2-2}{(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)}=\int \frac{ex+f}{x^2-\sqrt{3}x+1}+\frac{gx+h}{x^2+\sqrt{3}x+1}$$$${ex}^3+\sqrt{3}{ex}^2+ex+{fx}^2+\sqrt{3}fx+f+{gx}^3-\sqrt{3}{gx}^2+gx+{hx}^2-\sqrt{3}hx+h$$$$(e+g)x^3+(\sqrt{3}e+f-\sqrt{3}g+h)x^2+(e+\sqrt{3}f+g-\sqrt{3}h)x+(f+h)$$$$e+g=0\Rrightarrow \sqrt{3}(e-g)+f+h=1\Rrightarrow f+h=-2$$$$e+g+\sqrt{3}(f-h)=0\Rrightarrow f-h=0\Rrightarrow f=-1\Rrightarrow h=-1$$$$e-g=\sqrt{3}\Rrightarrow e=\frac{\sqrt{3}}{2}\Rrightarrow g=\frac{-\sqrt{3}}{2}$$$$\int \frac{\frac{\sqrt{3}}{2}x-1}{x^2-\sqrt{3}x+1}+\int \frac{\frac{-\sqrt{3}}{2}x-1}{x^2+\sqrt{3}x+1}$$$$\int \frac{\frac{\sqrt{3}}{4}.2x-\sqrt{3}.\frac{\sqrt{3}}{4}-\frac{1}{4}}{x^2-\sqrt{3}x+1}-\int \frac{\frac{\sqrt{3}}{4}.2x+\sqrt{3}.\frac{\sqrt{3}}{4}+\frac{1}{4}}{x^2+\sqrt{3}x+1}$$$$\frac{\sqrt{3}}{4}.ln\mid x^2-\sqrt{3}x+1\mid -\frac{\sqrt{3}}{4}.ln\mid x^2+\sqrt{3}x+1\mid -\frac{1}{4}\int \frac{1}{x^2-\sqrt{3}x+1}-\frac{1}{4}\int \frac{1}{x^2+\sqrt{3}x+1}$$$$\int \frac{1}{x^2-\sqrt{3}x+1}\Rrightarrow \int \frac{1}{{(x-\frac{\sqrt{3}}{2})}^2+\frac{1}{4}}\Rrightarrow 4\int \frac{1}{{(2x-\sqrt{3})}^2+1}$$$$2x-\sqrt{3}=v\Rrightarrow 2dx=dv$$$$2\int \frac{dv}{v^2+1}\Rrightarrow 2{tan}^{-1}\left(v\right)=2{tan}^{-1}(2x-\sqrt{3})$$$$\int \frac{1}{x^2+\sqrt{3}x+1}\Rrightarrow 4\int \frac{1}{{(2x+\sqrt{3})}^2+1}=2{tan}^{-1}(2x+\sqrt{3})$$$$\frac{1}{3}\int \frac{-x^2+2}{x^4-x^2+1}=\frac{\sqrt{3}}{12}.ln\mid \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}\mid +\frac{1}{6}(arctan(2x-\sqrt{3})+arctan(2x+\sqrt{3}))$$$$\int \frac{1}{x^6+1}=\frac{\sqrt{3}}{12}.ln\mid \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1}\mid +\frac{2arctan\left(x\right)+arctan(2x-\sqrt{3})+arctan(2x+\sqrt{3})}{6}$$

Answered by aba last updated on 16/Jun/23

$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^6+1}=\int \frac{{\mathrm{x}}^2+1-{\mathrm{x}}^2}{({\mathrm{x}}^2+1)({\mathrm{x}}^4-{\mathrm{x}}^2+1)}\mathrm{dx}$$$$=\int \frac{\cancel{{\mathrm{x}}^2+1}}{\cancel{({\mathrm{x}}^2+1)}({\mathrm{x}}^4-{\mathrm{x}}^2+1)}\mathrm{dx}-\int \frac{{\mathrm{x}}^2}{{\mathrm{x}}^6+1}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{({\mathrm{x}}^2+1)-({\mathrm{x}}^2-1)}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}\mathrm{dx}-\int \frac{{\mathrm{x}}^2}{{\left({\mathrm{x}}^3\right)}^2+1}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{1+\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2-1+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}-\frac{1}{2}\int \frac{1-\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2-1+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}-\int \frac{{\mathrm{x}}^2}{{\left({\mathrm{x}}^3\right)}^2+1}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{1+\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+1}\mathrm{dx}+\frac{1}{2}\int \frac{1-\frac{1}{{\mathrm{x}}^2}}{3-{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2}\mathrm{dx}-\int \frac{{\mathrm{x}}^2}{{\left({\mathrm{x}}^3\right)}^2+1}\mathrm{dx}$$$$=\frac{1}{2}\arctan (\mathrm{x}-\frac{1}{\mathrm{x}})+\frac{1}{2\sqrt{3}}\mathrm{arcot}\left(\frac{1}{\sqrt{3}}(\mathrm{x}+\frac{1}{\mathrm{x}})\right)-\frac{1}{3}\arctan \left({\mathrm{x}}^3\right)+\mathrm{k}$$$$$$

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