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Question Number 50384 by prof Abdo imad last updated on 16/Dec/18
find ∫ (dx/((1−x^2 )(1−x^3 ))) 2) calculate ∫_2 ^(√5) (dx/((1−x^2 )(1−x^3 )))
$${find}\:\int\:\:\:\frac{{dx}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)} \\ $$
Commented by Abdo msup. last updated on 23/Dec/18
let decompose F(x)=(1/((1−x^2 )(1−x^3 ))) F(x)=(1/((x−1)^2 (x+1)(x^2 +x+1))) =(a/(x−1)) +(b/((x−1)^2 )) +(c/(x+1)) +((dx +f)/(x^2 +x+1)) b=lim_(x→1) (x−1)^2 F(x)=(1/6) c =lim_(x→−1) (x+1)F(x)=(1/4) ⇒ F(x)=(a/(x−1)) +(1/(6(x−1)^2 )) +(1/(4(x+1)))+((dx+f)/(x^2 +x+1)) lim_(x→+∞) xF(x)=a +(1/4) +d=0 ⇒a+d=−(1/4) ⇒ F(x)= (a/(x−1)) +(1/(6(x−1)^2 )) +(1/(4(x+1))) +(((−a−(1/4))x +f)/(x^2 +x+1)) F(0)=1 =−a +(1/6) +(1/4) +f ⇒−a+f=1−(1/6)−(1/4) =(5/6)−(1/4) =((14)/(24)) =(7/(12)) F(2)=a +(1/6) +(1/(12)) +(((−2a−(1/2))+f)/7) = (1/(21)) ⇒ (1−(2/7))a +(1/4) −(1/(14)) +(f/7) =(1/(21)) ⇒ (5/7)a + ((10)/(56)) +(f/7) =(1/(21)) ⇒ 5a +((70)/(56)) +f =(1/3) ⇒ 5a +((35)/(28)) +f =(1/3) ⇒5a +f =(1/3)−((35)/(28)) be continued.....
$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$$=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{dx}\:+{f}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{6}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}+\frac{{dx}+{f}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)={a}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+{d}=\mathrm{0}\:\Rightarrow{a}+{d}=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{6}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}\:+\frac{\left(−{a}−\frac{\mathrm{1}}{\mathrm{4}}\right){x}\:+{f}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{1}\:=−{a}\:+\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+{f}\:\Rightarrow−{a}+{f}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{\mathrm{14}}{\mathrm{24}}\:=\frac{\mathrm{7}}{\mathrm{12}} \\ $$$${F}\left(\mathrm{2}\right)={a}\:+\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{12}}\:+\frac{\left(−\mathrm{2}{a}−\frac{\mathrm{1}}{\mathrm{2}}\right)+{f}}{\mathrm{7}}\:=\:\frac{\mathrm{1}}{\mathrm{21}}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{7}}\right){a}\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{14}}\:+\frac{{f}}{\mathrm{7}}\:=\frac{\mathrm{1}}{\mathrm{21}}\:\Rightarrow \\ $$$$\frac{\mathrm{5}}{\mathrm{7}}{a}\:+\:\frac{\mathrm{10}}{\mathrm{56}}\:+\frac{{f}}{\mathrm{7}}\:=\frac{\mathrm{1}}{\mathrm{21}}\:\Rightarrow\:\mathrm{5}{a}\:+\frac{\mathrm{70}}{\mathrm{56}}\:+{f}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{5}{a}\:+\frac{\mathrm{35}}{\mathrm{28}}\:+{f}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{5}{a}\:+{f}\:=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{35}}{\mathrm{28}} \\ $$$${be}\:{continued}….. \\ $$$$ \\ $$
Answered by ajfour last updated on 16/Dec/18
I = ∫(dx/((1−x^2 )(1−x^3 ))) Let (1/((1−x)^2 (1+x)(1+x+x^2 ))) = (A/((1−x)^2 ))+(B/(1−x))+(C/(1+x))+((Dx+E)/(1+x+x^2 )) A = (1/6) , C = (1/4) 1 = (1/6)(1+x)(1+x+x^2 ) +B(1+x)(1−x^3 ) +(1/4)(1−x)(1−x^3 ) +(Dx+E)(1−x)^2 (1+x) coeff. of x^4 = −B+(1/4)+D = 0 coeff. of x^3 = (1/6)−B−(1/4)+E−D=0 constant term coeff. = (1/6)+B+(1/4)+E = 1 ....
$${I}\:=\:\int\frac{{dx}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)}\: \\ $$$${Let}\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)}\: \\ $$$$\:\:\:\:\:=\:\frac{{A}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{{B}}{\mathrm{1}−{x}}+\frac{{C}}{\mathrm{1}+{x}}+\frac{{Dx}+{E}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} } \\ $$$${A}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:\:,\:\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+{B}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:+\left({Dx}+{E}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}+{x}\right) \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{4}} \:=\:−{B}+\frac{\mathrm{1}}{\mathrm{4}}+{D}\:=\:\mathrm{0} \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{6}}−{B}−\frac{\mathrm{1}}{\mathrm{4}}+{E}−{D}=\mathrm{0} \\ $$$${constant}\:{term}\:{coeff}.\: \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{6}}+{B}+\frac{\mathrm{1}}{\mathrm{4}}+{E}\:=\:\mathrm{1} \\ $$$$…. \\ $$

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