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Question Number 69894 by Rio Michael last updated on 28/Sep/19
∫ ((2x^5 −x)/(x^3 −2))dx
$$\int\:\frac{\mathrm{2}{x}^{\mathrm{5}} −{x}}{{x}^{\mathrm{3}} −\mathrm{2}}{dx} \\ $$
Commented by abdo mathsup 649 cc last updated on 29/Sep/19
I=∫ ((2x^5 −x)/(x^3 −2))dx =∫ ((2x^2 (x^3 −2)+4x^2 −x)/(x^3 −2))dx =∫ 2x^2 dx +∫ ((4x^2 −x)/(x^3 −2))dx ∫ 2x^(2 ) =(2/3)x^(3 ) +c let decompose F(x)=((4x^2 −x)/(x^3 −2)) ⇒F(x)=((4x^2 −x)/((x−(^3 (√2))(x^2 +x(^3 (√2))+(^3 (√2))^2 ))) let put^3 (√2)=α ⇒F(x)=((4x^2 −x)/((x−α)(x^2 +αx +α^2 ))) =(a/(x−α)) +((bx +c)/(x^2 +αx +α^2 )) a =((4α^2 −α)/(3α^2 )) =((4α−1)/(3α))=(4/3)−(1/(3α)) lim_(x−→+∞) xF(x)=4 =a+b ⇒b=4−a =4−((4/3)−(1/(3α)))=(8/3) +(1/(3α)) F(0)=0=−(a/α) +(c/α^2 ) =((c−αa)/α^2 ) ⇒c=αa=((4α)/3)−(1/3) ∫ F(x)dx =aln∣x−α∣ +(1/2)∫ ((2bx+2c)/(x^2 +αx +α^2 ))dx =aln∣x−α∣+(b/2) ∫ ((2x+α−α +2c)/(x^2 +αx +α^2 ))dx =aln∣x−α∣ +(b/2)ln(x^2 +αx +α^2 )+(((2c−α)b)/2)∫(dx/(x^2 +αx +α^2 )) and ∫ (dx/(x^2 +αx+α^2 ))=∫ (dx/(x^2 +2(α/2)x +(α^2 /4)+α^2 −(α^2 /4))) =∫ (dx/((x+(α/2))^2 +((3α^2 )/4))) =_(x+(α/2)=((√3)/2)u) =(4/(3α^2 )) ∫ (1/(1+u^2 ))((√3)/2)du =(2/( (√3)α^2 ))arctan(((2x+α)/( (√3)))) +c ⇒ I =(2/3)x^3 +aln∣x−α∣+(b/2)ln(x^2 +αx +α^2 ) +(((2c−α)b)/( (√3)α^2 )) arctan(((2x+α)/( (√3)))) +C with α=^3 (√2)
$${I}=\int\:\:\frac{\mathrm{2}{x}^{\mathrm{5}} −{x}}{{x}^{\mathrm{3}} −\mathrm{2}}{dx}\:=\int\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{2}\right)+\mathrm{4}{x}^{\mathrm{2}} −{x}}{{x}^{\mathrm{3}} −\mathrm{2}}{dx} \\ $$$$=\int\:\:\mathrm{2}{x}^{\mathrm{2}} {dx}\:+\int\:\:\frac{\mathrm{4}{x}^{\mathrm{2}} −{x}}{{x}^{\mathrm{3}} −\mathrm{2}}{dx} \\ $$$$\int\:\mathrm{2}{x}^{\mathrm{2}\:} \:=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}\:} \:+{c}\:\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} −{x}}{{x}^{\mathrm{3}} −\mathrm{2}}\:\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} −{x}}{\left({x}−\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +{x}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)+\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)^{\mathrm{2}} \right)\right.} \\ $$$${let}\:{put}\:^{\mathrm{3}} \sqrt{\mathrm{2}}=\alpha\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} −{x}}{\left({x}−\alpha\right)\left({x}^{\mathrm{2}} +\alpha{x}\:+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{{a}}{{x}−\alpha}\:+\frac{{bx}\:+{c}}{{x}^{\mathrm{2}} +\alpha{x}\:+\alpha^{\mathrm{2}} } \\ $$$${a}\:=\frac{\mathrm{4}\alpha^{\mathrm{2}} −\alpha}{\mathrm{3}\alpha^{\mathrm{2}} }\:=\frac{\mathrm{4}\alpha−\mathrm{1}}{\mathrm{3}\alpha}=\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}\alpha} \\ $$$${lim}_{{x}−\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{4}\:={a}+{b}\:\Rightarrow{b}=\mathrm{4}−{a} \\ $$$$=\mathrm{4}−\left(\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}\alpha}\right)=\frac{\mathrm{8}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}\alpha} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}=−\frac{{a}}{\alpha}\:+\frac{{c}}{\alpha^{\mathrm{2}} }\:=\frac{{c}−\alpha{a}}{\alpha^{\mathrm{2}} }\:\Rightarrow{c}=\alpha{a}=\frac{\mathrm{4}\alpha}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\int\:\:{F}\left({x}\right){dx}\:={aln}\mid{x}−\alpha\mid\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{bx}+\mathrm{2}{c}}{{x}^{\mathrm{2}} \:+\alpha{x}\:+\alpha^{\mathrm{2}} }{dx} \\ $$$$={aln}\mid{x}−\alpha\mid+\frac{{b}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{2}{x}+\alpha−\alpha\:+\mathrm{2}{c}}{{x}^{\mathrm{2}} \:+\alpha{x}\:+\alpha^{\mathrm{2}} }{dx} \\ $$$$={aln}\mid{x}−\alpha\mid\:+\frac{{b}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+\alpha{x}\:+\alpha^{\mathrm{2}} \right)+\frac{\left(\mathrm{2}{c}−\alpha\right){b}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} \:+\alpha{x}\:+\alpha^{\mathrm{2}} } \\ $$$${and}\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\alpha{x}+\alpha^{\mathrm{2}} }=\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}\frac{\alpha}{\mathrm{2}}{x}\:+\frac{\alpha^{\mathrm{2}} }{\mathrm{4}}+\alpha^{\mathrm{2}} −\frac{\alpha^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$=\int\:\:\frac{{dx}}{\left({x}+\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{4}}}\:=_{{x}+\frac{\alpha}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}\alpha^{\mathrm{2}} }\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}\alpha^{\mathrm{2}} }{arctan}\left(\frac{\mathrm{2}{x}+\alpha}{\:\sqrt{\mathrm{3}}}\right)\:+{c}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:+{aln}\mid{x}−\alpha\mid+\frac{{b}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+\alpha{x}\:+\alpha^{\mathrm{2}} \right) \\ $$$$+\frac{\left(\mathrm{2}{c}−\alpha\right){b}}{\:\sqrt{\mathrm{3}}\alpha^{\mathrm{2}} }\:{arctan}\left(\frac{\mathrm{2}{x}+\alpha}{\:\sqrt{\mathrm{3}}}\right)\:+{C}\:\:{with}\:\alpha=^{\mathrm{3}} \sqrt{\mathrm{2}} \\ $$
Commented by Rio Michael last updated on 29/Sep/19
thanks sir
$${thanks}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 29/Sep/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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