Question Number 86491 by john santu last updated on 29/Mar/20
$$\int\:\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}\:\mathrm{dx}\:?\: \\ $$
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
$$\int\frac{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} }}{\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}}}\right)\:+{C} \\ $$
Commented by john santu last updated on 29/Mar/20
$$\mathrm{good}\:\mathrm{answer} \\ $$
Commented by som(math1967) last updated on 29/Mar/20
$${Thank}\:{you}\:{sir} \\ $$
Answered by MJS last updated on 29/Mar/20
$$\mathrm{other}\:\mathrm{method}: \\ $$$$\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right)+{C} \\ $$
Commented by john santu last updated on 29/Mar/20
$$\mathrm{ostrogradetski}? \\ $$
Commented by MJS last updated on 29/Mar/20
$$\mathrm{no}.\:\mathrm{just}\:\mathrm{decomposing} \\ $$