Question Number 105411 by Ar Brandon last updated on 28/Jul/20
$$\int\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}^{\mathrm{6}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{x} \\ $$$${I}\mathrm{s}\:{there}\:{any}\:{special}\:{method}\:{of}\:{decomposition} \\ $$$${other}\:{than}\:{the}\:{use}\:{of}\:{partial}\:{fractions}\:? \\ $$
Answered by Dwaipayan Shikari last updated on 28/Jul/20
$$\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{6}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{2}}{{x}^{\mathrm{6}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$−\frac{\mathrm{3}}{\mathrm{5}{x}^{\mathrm{5}} }+\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{3}} }−\frac{\mathrm{2}}{{x}}−\mathrm{2}{tan}^{−\mathrm{1}} {x}+{C} \\ $$
Commented by Dwaipayan Shikari last updated on 28/Jul/20
$${I}\:{think}\:{so}\:.\:{Is}\:{it}\:{a}\:{better}\:{way}\:{that}\:{you}\:{have}\:{mentioned}? \\ $$
Commented by Dwaipayan Shikari last updated on 28/Jul/20
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Commented by Ar Brandon last updated on 28/Jul/20
Perfect, bro. Exactly what I needed. Thanks��
Commented by Ar Brandon last updated on 28/Jul/20
$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{x}^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:,\: \\ $$