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1-x-6-x-3-dx-

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Question Number 64227 by Tony Lin last updated on 16/Jul/19
∫(1/(x^6 +x^3 ))dx=?
$$\int\frac{\mathrm{1}}{{x}^{\mathrm{6}} +{x}^{\mathrm{3}} }{dx}=? \\ $$
Answered by ajfour last updated on 16/Jul/19
I=∫(dx/(x^3 (1+x^3 ))) (1/(x^3 (1+x^3 )))=(A/x)+(B/x^2 )+(C/x^3 )+(D/(1+x))+((Ex+F)/(x^2 −x+1)) ⇒ 1=(1+x^3 )(Ax^2 +Bx+C) +Dx^3 (x^2 −x+1)+(Ex+F)x^3 (1+x) x=0 ⇒ C=1 x=−1 ⇒ D=−(1/3) ⇒ A+D+E=0 B−D+E+F =0 C+D+F =0 A=0, B=0 ⇒ E=(1/3), F=−(2/3) ⇒ I=∫ (dx/x^3 )−(1/3)∫(dx/(1+x))+(1/3)∫ (((x−2)dx)/(x^2 −x+1)) =−(1/(2x^2 ))−(1/3)ln (1+x)+(1/6)ln (x^2 −x+1) −∫(dx/((x−(1/2))^2 +(((√3)/2))^2 )) I=−(1/(2x^2 ))−(1/3)ln (1+x)+(1/6)ln (x^2 −x+1) −(2/( (√3)))tan^(−1) (((2x−1)/( (√3))))+c .
$${I}=\int\frac{{dx}}{{x}^{\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{3}} \right)} \\ $$$$\:\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{3}} \right)}=\frac{{A}}{{x}}+\frac{{B}}{{x}^{\mathrm{2}} }+\frac{{C}}{{x}^{\mathrm{3}} }+\frac{{D}}{\mathrm{1}+{x}}+\frac{{Ex}+{F}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{1}=\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\left({Ax}^{\mathrm{2}} +{Bx}+{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+{Dx}^{\mathrm{3}} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\left({Ex}+{F}\right){x}^{\mathrm{3}} \left(\mathrm{1}+{x}\right) \\ $$$$\:{x}=\mathrm{0}\:\Rightarrow\:{C}=\mathrm{1}\: \\ $$$$\:{x}=−\mathrm{1}\:\Rightarrow\:{D}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:{A}+{D}+{E}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{B}−{D}+{E}+{F}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{C}+{D}+{F}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{A}=\mathrm{0},\:{B}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{E}=\frac{\mathrm{1}}{\mathrm{3}},\:{F}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\:{I}=\int\:\frac{{dx}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{\left({x}−\mathrm{2}\right){dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{1}+{x}\right)+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:−\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{1}+{x}\right)+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c}\:. \\ $$
Commented by Tony Lin last updated on 16/Jul/19
thanks sir
$${thanks}\:{sir} \\ $$

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