Integration Questions

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

$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{x}^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:,\: \\$$