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Question Number 166939 by cortano1 last updated on 03/Mar/22

Answered by som(math1967) last updated on 03/Mar/22

$$\int \frac{{(x^2+1)}^2-2x^2}{x^6+1}dx$$$$\int \frac{{(x^2+1)}^2}{(x^2+1)(x^4-x^2+1)}dx-2\int \frac{x^{2}dx}{{\left(x^3\right)}^2+1}$$$$=\int \frac{(x^2+1)dx}{(x^4-x^2+1)}-\frac{2}{3}\int \frac{3x^2}{{\left(x^3\right)}^2+1}dx$$$$=\int \frac{(1+\frac{1}{x^2})dx}{x^2+\frac{1}{x^2}-1}-\frac{2}{3}\int \frac{d\left(x^3\right)}{{\left(x^3\right)}^2+1}$$$$=\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+1}-\frac{2}{3}{\tan }^{-1}{x}^3$$$$={\tan }^{-1}(x-\frac{1}{x})-\frac{2}{3}{\tan }^{-1}{x}^3+c$$

Commented by MJS_new last updated on 03/Mar/22

$$\mathrm{yes}!$$$$\mathrm{my}\mathrm{path}:$$$$\int \frac{x^4+1}{x^6+1}dx=$$$$=\frac{1}{6}\int \frac{dx}{x^2-\sqrt{3}x+1}+\frac{1}{6}\int \frac{dx}{x^2+\sqrt{3}x+1}+\frac{2}{3}\int \frac{dx}{x^2+1}=$$$$=\frac{1}{3}\arctan (2x-\sqrt{3})+\frac{1}{3}\arctan (2x+\sqrt{3})+\frac{2}{3}\arctan x+C$$

Commented by som(math1967) last updated on 04/Mar/22

$$nicepathsir$$

Commented by MJS_new last updated on 04/Mar/22

$$\mathrm{I}\mathrm{like}\mathrm{yours}\mathrm{more},\mathrm{because}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{obvious}$$

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