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Question Number 105411 by Ar Brandon last updated on 28/Jul/20

∫((x^2 +3)/(x^6 (x^2 +1)))dx Is there any special method of decomposition other than the use of partial fractions ?

$$\int\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}^{\mathrm{6}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{d}{x} \\ $$$${I}\mathrm{s}\:{there}\:{any}\:{special}\:{method}\:{of}\:{decomposition} \\ $$$${other}\:{than}\:{the}\:{use}\:{of}\:{partial}\:{fractions}\:? \\ $$

Answered by Dwaipayan Shikari last updated on 28/Jul/20

∫((x^2 +1)/(x^6 (x^2 +1)))+(2/(x^6 (x^2 +1)))dx ∫(1/x^6 )+2∫(1/x^4 )((1/x^2 )−(1/(x^2 +1)))dx 3∫(1/x^6 )−2∫(1/(x^4 (x^2 +1)))dx 3∫(1/x^6 )−2∫(1/x^2 )((1/x^2 )−(1/(x^2 +1)))dx 3∫(1/x^6 )−2∫(1/x^4 )+2∫(1/(x^2 (x^2 +1)))dx 3∫(1/x^6 )−2∫(1/x^4 )+2∫(1/x^2 )−2∫(1/(x^2 +1))dx −(3/(5x^5 ))+(2/(3x^3 ))−(2/x)−2tan^(−1) x+C

$$\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{6}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{2}}{{x}^{\mathrm{6}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\mathrm{3}\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$−\frac{\mathrm{3}}{\mathrm{5}{x}^{\mathrm{5}} }+\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{3}} }−\frac{\mathrm{2}}{{x}}−\mathrm{2}{tan}^{−\mathrm{1}} {x}+{C} \\ $$

Commented by Dwaipayan Shikari last updated on 28/Jul/20

I think so . Is it a better way that you have mentioned?

$${I}\:{think}\:{so}\:.\:{Is}\:{it}\:{a}\:{better}\:{way}\:{that}\:{you}\:{have}\:{mentioned}? \\ $$

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

(1/(x^2 (x^2 +1)))=((x^2 +1−x^2 )/(x^2 (x^2 +1)))=((x^2 +1)/(x^2 (x^2 +1)))−(x^2 /(x^2 (x^2 +1))) =(1/x^2 )−(1/(x^2 +1)) ,

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

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