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Question Number 86491 by john santu last updated on 29/Mar/20

$$\int \frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^4+1}\mathrm{dx}?$$

Commented by john santu last updated on 29/Mar/20

$$\mathrm{dear}\mathrm{prof}\mathrm{mr}\mathrm{mjs}.\mathrm{what}\mathrm{the}\mathrm{super}$$$$\mathrm{easy}\mathrm{method}?$$

Answered by som(math1967) last updated on 29/Mar/20

$$\int \frac{\frac{x^2+1}{x^2}}{\frac{x^4+1}{x^2}}dx$$$$\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$$$$\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+{\left(\sqrt{2}\right)}^2}$$$$\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C$$

Commented by john santu last updated on 29/Mar/20

$$\mathrm{good}\mathrm{answer}$$

Commented by som(math1967) last updated on 29/Mar/20

$$Thankyousir$$

Answered by MJS last updated on 29/Mar/20

$$\mathrm{other}\mathrm{method}:$$$$\int \frac{x^2+1}{x^4+1}dx=$$$$=\frac{1}{2}\int \frac{dx}{x^2-\sqrt{2}x+1}+\frac{1}{2}\int \frac{dx}{x^2+\sqrt{2}x+1}=$$$$=\frac{\sqrt{2}}{2}(\arctan (\sqrt{2}x-1)+\arctan (\sqrt{2}x+1))+C$$

Commented by john santu last updated on 29/Mar/20

$$\mathrm{ostrogradetski}?$$

Commented by MJS last updated on 29/Mar/20

$$\mathrm{no}.\mathrm{just}\mathrm{decomposing}$$

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