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Question Number 76053 by john santuy last updated on 23/Dec/19

$$howIcalculate\int \frac{1}{x^8+x^2}dx?$$

Commented by abdomathmax last updated on 23/Dec/19

$$decomposethefractionF\left(x\right)=\frac{1}{x^2+x^8}andsee$$$$thatF\left(x\right)=\frac{1}{x^2(x^6+1)}=\frac{1}{x^2({\left(x^2\right)}^3+1)}$$$$=\frac{1}{x^2(x^2+1)(x^4-x^2+1)}=....$$

Commented by benjo last updated on 23/Dec/19

$$\mathrm{thanks}\mathrm{sir}$$

Answered by MJS last updated on 23/Dec/19

$$\frac{1}{x^8+x^2}=\frac{1}{x^2(x^6+1)}=$$$$=\frac{1}{x^2}-\frac{1}{3(x^2+1)}+\frac{\sqrt{3}x+1}{6(x^2+\sqrt{3}x+1)}-\frac{\sqrt{3}x-1}{6(x^2-\sqrt{3}x+1)}$$$$\mathrm{now}\mathrm{solve}\mathrm{these}$$

Commented by benjo last updated on 23/Dec/19

$$\mathrm{thanks}\mathrm{sir}]$$

Commented by vishalbhardwaj last updated on 23/Dec/19

$$\mathrm{sir}\mathrm{this}\mathrm{procedure}\mathrm{is}\mathrm{very}\mathrm{difficult}\mathrm{to}\mathrm{undersatand}$$$$,\mathrm{please}\mathrm{explain}\mathrm{this}\mathrm{once}\mathrm{again}\mathrm{when}\mathrm{you}\mathrm{are}\mathrm{direct}$$$$\mathrm{ly}\mathrm{taken}\mathrm{any}\mathrm{value}\mathrm{of}\mathrm{x}$$$$$$

Commented by MJS last updated on 23/Dec/19

$$\mathrm{the}\mathrm{decomposing}\mathrm{or}\mathrm{the}\mathrm{integration}?$$

Commented by vishalbhardwaj last updated on 23/Dec/19

$$\mathrm{decomposing}$$

Commented by MJS last updated on 24/Dec/19

$$x^8+x^2=x^2(x^6+1)=$$$$=x^2({\left(x^2\right)}^3+1)=$$$$\left[\begin{array}{c}{p}^3+1=(p+1)(p^2-p+1)\end{array}\right]$$$$=x^2(x^2+1)(x^4-x^2+1)=$$$$\left[\begin{array}{c}{x}^4-x^2+1=0\mathrm{let}x=\sqrt{q}\\ q^2-q+1=0\Rightarrow q=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}\\ \mathrm{these}\mathrm{are}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{and}{3}^{\mathrm{rd}}\mathrm{solutions}\\ \mathrm{of}{q}^3+1=0\Rightarrow q={\mathrm{e}}^{\pm \mathrm{i}\frac{\pi }{3}}\\ \Rightarrow x={\mathrm{e}}^{\pm \mathrm{i}\frac{\pi }{6}}\vee x={\mathrm{e}}^{\pm \mathrm{i}\frac{5\pi }{6}}\\ \Rightarrow x^4-x^2+1=\\ =(x-{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{6}})(x-{\mathrm{e}}^{\mathrm{i}\frac{\pi }{6}})(x-{\mathrm{e}}^{-\mathrm{i}\frac{5\pi }{6}})(x-{\mathrm{e}}^{\mathrm{i}\frac{5\pi }{6}})=\\ =(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)\end{array}\right]$$$$=x^2(x^2+1)(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)$$$$\Rightarrow$$$$\frac{1}{x^8+x^2}=\frac{ax+b}{x^2}+\frac{cx+d}{x^2+1}+\frac{ex+f}{x^2-\sqrt{3}x+1}+\frac{gx+h}{x^2+\sqrt{3}x+1}$$$$\mathrm{now}\mathrm{use}\mathrm{your}\left(\mathrm{hopefully}\right)\mathrm{learned}\mathrm{method}$$

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