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Question Number 86824 by Ar Brandon last updated on 31/Mar/20

$$\int \frac{x^6+x^2}{x^8-x^4+1}dx$$

Answered by TANMAY PANACEA. last updated on 31/Mar/20

$$\int \frac{x^2+\frac{1}{x^2}}{x^4-1+\frac{1}{x^4}}dx$$$$\int \frac{x^2+\frac{1}{x^2}}{{(x^2+\frac{1}{x^2})}^2-3}dx$$$$\frac{1}{2}\int \frac{x^2+\frac{1}{x^2}+\sqrt{3}+x^2+\frac{1}{x^2}-\sqrt{3}}{(x^2+\frac{1}{x^2}+\sqrt{3})(x^2+\frac{1}{x^2}-\sqrt{3})}dx$$$$\frac{1}{2}\int \frac{dx}{x^2+\frac{1}{x^2}-\sqrt{3}}+\frac{1}{2}\int \frac{dx}{x^2+\frac{1}{x^2}+\sqrt{3}}$$$$\frac{1}{4}\int \frac{1-\frac{1}{x^2}+1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-\sqrt{3}}dx+\frac{1}{4}\int \frac{1-\frac{1}{x^2}+1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+\sqrt{3}}dx$$$$\frac{1}{4}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-(2+\sqrt{3})}+\frac{1}{4}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2-\sqrt{3}}+\frac{1}{4}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-2+\sqrt{3}}+\frac{1}{4}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2+\sqrt{3}}$$$$\frac{1}{4}\times \frac{1}{2\sqrt{2+\sqrt{3}}}ln\left(\frac{x+\frac{1}{x}-\sqrt{2+\sqrt{3}}}{x+\frac{1}{x}+\sqrt{2+\sqrt{3}}}\right)=I_1$$$$\frac{1}{4}\times \frac{1}{\sqrt{2-\sqrt{3}}}{tan}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2-\sqrt{3}}}\right)=I_2$$$$\frac{1}{4}\times \frac{1}{2\sqrt{2-\sqrt{3}}}ln\left(\frac{x+\frac{1}{x}-\sqrt{2-\sqrt{3}}}{x+\frac{1}{x}+\sqrt{2-\sqrt{3}}}\right)=I_3$$$$\frac{1}{4}\times \frac{1}{\sqrt{2+\sqrt{3}}}{tan}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2+\sqrt{3}}}\right)$$$$\mathit{p}\mathit{l}\mathit{s}\mathit{a}\mathit{d}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{m}...$$

Commented by Ar Brandon last updated on 02/Apr/20

$$Amazing!$$

Commented by TANMAY PANACEA. last updated on 02/Apr/20

$$thankyousir...$$

Answered by M±th+et£s last updated on 03/Apr/20

$$\frac{1}{2}\int \frac{2(x^6+x^2)}{x^8+2x^4+1-3x^4}dx=\frac{1}{2}\int \frac{x^6+x^6+\sqrt{3}{x}^4-\sqrt{3}{x}^4+x^2+x^2}{{(x^4+1)}^2-3x^4}dx$$$$\frac{1}{2}\int \frac{x^6+\sqrt{3}{x}^4+x^2+x^6-\sqrt{3}{x}^4+x^2}{(x^4+1-\sqrt{3}{x}^2)(x^4+1+\sqrt{3}{x}^2)}dx$$$$\frac{1}{2}\int \frac{x^2(x^4+\sqrt{3}{x}^2+1)+x^2(x^4-\sqrt{3}{x}^2+1)}{(x^4+1-\sqrt{3}{x}^2)(x^4+1+\sqrt{3}{x}^2)}dx$$$$\frac{1}{4}\int (\frac{2x^2}{(x^4+1+\sqrt{3}{x}^2)}+\frac{2x^2}{(x^4+1-\sqrt{3}{x}^2})dx$$$$\frac{1}{4}\int (\frac{x^2+1+x^2-1}{(x^4+1+\sqrt{3}{x}^2)}+\frac{x^2+1+x^2-1}{(x^4+1-\sqrt{3}{x}^2)})dx$$$$\frac{1}{4}\int (\frac{x^2+1}{x^4+1+\sqrt{3}{x}^2}+\frac{x^2-1}{x^4+1+\sqrt{3}{x}^2}+\frac{x^2+1}{x^4+1-\sqrt{3}{x}^2}+\frac{x^2-1}{x^4+1-\sqrt{3}{x}^2})dx$$$$\frac{1}{4}\int (\frac{1+1/x^2}{x^2+\sqrt{3}+1/x^2}dx+\frac{1-1/x^2}{x^2+\sqrt{3}+1/x^2}+\frac{1+1/x^2}{x^2-\sqrt{3}+1/x^2}+\frac{1-1/x^2}{x^2-\sqrt{3}+1/x^2})dx$$$$\frac{1}{4}\int (\frac{1+1/x^2}{x^2-2+1/x^2-\sqrt{3}+2}+\frac{1-1/x^2}{x^2+2+1/x^2-2-\sqrt{3}}+\frac{1+1/x^2}{x^2-2+1/x^2+2+\sqrt{3}}+\frac{1-1/x^2}{x^2+2+1/x^2-2+\sqrt{3}})$$$$\frac{1}{4}\int (\frac{1+1/x^2}{{(x-1/x)}^2-(\sqrt{3}-2)}+\frac{1-1/x^2}{{(x+1/x)}^2-(2+\sqrt{3})}+\frac{1+1/x^2}{{(x-1/x)}^2+(\sqrt{3}+2)}+\frac{1-1/x^2}{{(x-1/x)}^2+(\sqrt{3}-2)})dx$$$$\frac{1}{4\sqrt{2-\sqrt{3}}}{tanh}^{-1}\frac{1}{\sqrt{2-\sqrt{3}}}(x-\frac{1}{x})+\frac{1}{4\sqrt{2-\sqrt{3}}}{tanh}^{-1}\frac{1}{\sqrt{2-\sqrt{3}}}(x+\frac{1}{x})+\frac{1}{4\sqrt{\sqrt{3}+2}}{tan}^{-1}\frac{1}{\sqrt{\sqrt{3}+2}}(x-\frac{1}{x})+\frac{1}{4\sqrt{\sqrt{3}-2}}{tan}^{-1}\frac{1}{\sqrt{\sqrt{3}-2}}(x+\frac{1}{x})+c$$$$$$$$$$$$$$$$plscheck$$$$$$

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