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Question Number 73147 by amaramariche last updated on 06/Nov/19
∫((x−6)/(x^3 +1))dx
$$\int\frac{{x}−\mathrm{6}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 07/Nov/19
let A =∫ ((x−6)/(x^3 +1))dx let decompose F(x)=((x−6)/(x^3 +1))=((x−6)/((x+1)(x^2 −x+1))) F(x)=(a/(x+1)) +((bx+c)/(x^2 −x+1)) a=lim_(x→−1) (x+1)F(x)=((−7)/3) lim_(x→+∞) xF(x)=0=a+b ⇒b=(7/3) F(o)=−6 =a+c ⇒c=−6+(7/3) =−((11)/3) ⇒F(x)=((−7)/(3(x+1)))+(1/3)×((7x−11)/(x^2 −x+1)) ⇒A =∫ F(x)dx =−(7/3)ln∣x+1∣+(7/6)∫ ((2x−1+1)/(x^2 −x+1))−((11)/3)∫(dx/(x^2 −x+1)) =−(7/3)ln∣x+1∣+(7/6)ln∣x^2 −x+1∣−((15)/6) ∫ (dx/(x^2 −x+1)) ∫ (dx/(x^2 −x+1)) =∫ (dx/(x^2 −2(x/2)+(1/4)+1−(1/4))) =∫ (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)u) (4/3)∫ (1/(u^2 +1))((√3)/2)du =(2/( (√3))) arctan(((2x−1)/( (√3)))) ⇒ A =−(7/3)ln∣x+1∣+(7/6)ln∣x^2 −x+1∣−(5/( (√3))) arctan(((2x−1)/( (√3))))+C.
$${let}\:{A}\:=\int\:\frac{{x}−\mathrm{6}}{{x}^{\mathrm{3}} \:+\mathrm{1}}{dx}\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{{x}−\mathrm{6}}{{x}^{\mathrm{3}} \:+\mathrm{1}}=\frac{{x}−\mathrm{6}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${a}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{−\mathrm{7}}{\mathrm{3}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}={a}+{b}\:\Rightarrow{b}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$${F}\left({o}\right)=−\mathrm{6}\:={a}+{c}\:\Rightarrow{c}=−\mathrm{6}+\frac{\mathrm{7}}{\mathrm{3}}\:=−\frac{\mathrm{11}}{\mathrm{3}}\:\Rightarrow{F}\left({x}\right)=\frac{−\mathrm{7}}{\mathrm{3}\left({x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{7}{x}−\mathrm{11}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\Rightarrow{A}\:=\int\:{F}\left({x}\right){dx}\:=−\frac{\mathrm{7}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{7}}{\mathrm{6}}\int\:\:\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−\frac{\mathrm{11}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{7}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{7}}{\mathrm{6}}{ln}\mid{x}^{\mathrm{2}} −{x}+\mathrm{1}\mid−\frac{\mathrm{15}}{\mathrm{6}}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:=\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}\:=\int\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \:\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\int\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\Rightarrow \\ $$$${A}\:=−\frac{\mathrm{7}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{7}}{\mathrm{6}}{ln}\mid{x}^{\mathrm{2}} −{x}+\mathrm{1}\mid−\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C}. \\ $$
Answered by MJS last updated on 06/Nov/19
∫((x−6)/(x^3 +1))dx=−(7/3)∫(dx/(x+1))+(1/3)∫((7x−11)/(x^2 −x+1))dx= =−(7/3)ln (x+1) +(7/6)ln (x^2 −x+1) −((5(√3))/3)arctan (((2x−1)(√3))/3) = =(7/6)ln ((x^2 −x+1)/((x+1)^3 )) −((5(√3))/3)arctan (((2x−1)(√3))/3) +C
$$\int\frac{{x}−\mathrm{6}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx}=−\frac{\mathrm{7}}{\mathrm{3}}\int\frac{{dx}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{7}{x}−\mathrm{11}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$=−\frac{\mathrm{7}}{\mathrm{3}}\mathrm{ln}\:\left({x}+\mathrm{1}\right)\:+\frac{\mathrm{7}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:−\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\:= \\ $$$$=\frac{\mathrm{7}}{\mathrm{6}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\:+{C} \\ $$
Commented by mind is power last updated on 06/Nov/19
3 lign! nice sir
$$\mathrm{3}\:\mathrm{lign}!\:\mathrm{nice}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 06/Nov/19
thank you I′m just an integral lover, you know far more than me
$$\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{just}\:\mathrm{an}\:\mathrm{integral}\:\mathrm{lover},\:\mathrm{you}\:\mathrm{know}\:\mathrm{far}\:\mathrm{more} \\ $$$$\mathrm{than}\:\mathrm{me} \\ $$
Commented by mind is power last updated on 06/Nov/19
my bee im just older ,you are inteligent and you have creative solution sir im not genius ir somthing like/that[i worck hard evrey Day im just a hard worcker s
$$\mathrm{my}\:\mathrm{bee}\:\mathrm{im}\:\mathrm{just}\:\mathrm{older}\:,\mathrm{you}\:\mathrm{are}\:\mathrm{inteligent}\:\mathrm{and}\:\mathrm{you}\:\mathrm{have}\:\mathrm{creative}\:\mathrm{solution}\:\mathrm{sir} \\ $$$$\mathrm{im}\:\mathrm{not}\:\mathrm{genius}\:\mathrm{ir}\:\mathrm{somthing}\:\mathrm{like}/\mathrm{that}\left[\mathrm{i}\:\mathrm{worck}\:\mathrm{hard}\:\mathrm{evrey}\:\mathrm{Day}\right. \\ $$$$\mathrm{im}\:\mathrm{just}\:\mathrm{a}\:\mathrm{hard}\:\mathrm{worcker}\:\mathrm{s} \\ $$$$ \\ $$
Commented by MJS last updated on 07/Nov/19
haha! I studied decades ago but never made any grade, instead I became a musician creative yes, but not firm in many things... my age? my age in the year y_1 was a_1 my age in tbe year y_2 would be a_2 (but noone alive today will see that year) (1) a_1 ∣y_1 ∧a_2 ∣y_2 ∧a_1 ∣a_2 (2) (y_1 /a_1 )−1=a_2 ∧(y_2 /a_2 )−1=a_1 not sure if this is unique...
$$\mathrm{haha}!\:\mathrm{I}\:\mathrm{studied}\:\mathrm{decades}\:\mathrm{ago}\:\mathrm{but}\:\mathrm{never}\:\mathrm{made} \\ $$$$\mathrm{any}\:\mathrm{grade},\:\mathrm{instead}\:\mathrm{I}\:\mathrm{became}\:\mathrm{a}\:\mathrm{musician} \\ $$$$\mathrm{creative}\:\mathrm{yes},\:\mathrm{but}\:\mathrm{not}\:\mathrm{firm}\:\mathrm{in}\:\mathrm{many}\:\mathrm{things}… \\ $$$$\mathrm{my}\:\mathrm{age}? \\ $$$$\mathrm{my}\:\mathrm{age}\:\mathrm{in}\:\mathrm{the}\:\mathrm{year}\:{y}_{\mathrm{1}} \:\mathrm{was}\:{a}_{\mathrm{1}} \\ $$$$\mathrm{my}\:\mathrm{age}\:\mathrm{in}\:\mathrm{tbe}\:\mathrm{year}\:{y}_{\mathrm{2}} \:\mathrm{would}\:\mathrm{be}\:{a}_{\mathrm{2}} \:\left(\mathrm{but}\:\mathrm{noone}\right. \\ $$$$\left.\mathrm{alive}\:\mathrm{today}\:\mathrm{will}\:\mathrm{see}\:\mathrm{that}\:\mathrm{year}\right) \\ $$$$\left(\mathrm{1}\right)\:\:{a}_{\mathrm{1}} \mid{y}_{\mathrm{1}} \wedge{a}_{\mathrm{2}} \mid{y}_{\mathrm{2}} \wedge{a}_{\mathrm{1}} \mid{a}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\:\frac{{y}_{\mathrm{1}} }{{a}_{\mathrm{1}} }−\mathrm{1}={a}_{\mathrm{2}} \wedge\frac{{y}_{\mathrm{2}} }{{a}_{\mathrm{2}} }−\mathrm{1}={a}_{\mathrm{1}} \\ $$$$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{unique}… \\ $$
Commented by Rasheed.Sindhi last updated on 07/Nov/19
(1) a_1 ∣y_1 ∧a_2 ∣y_2 ∧a_1 ∣a_2 (2) (y_1 /a_1 )−1=a_2 ∧(y_2 /a_2 )−1=a_1 y_2 −y_1 =a_2 −a_1 >100 y_1 <2019 y_1 =a_1 k_(1 ) ,y_2 =a_2 k_2 ,a_2 =a_1 k_3 (Say) ( k_1 ,k_2 ,k_3 ∈N) y_1 =a_1 k_(1 ) ,y_2 =(a_1 k_3 )k_2 =a_1 k_2 k_3 y_2 −y_1 =a_2 −a_1 >100 (y_2 is very high) ⇒a_1 k_2 k_3 −a_1 k_(1 ) =a_1 k_3 −a_1 ⇒k_2 k_3 −k_(1 ) =k_3 −1 ⇒k_3 (k_2 −1)=k_1 −1 k_3 =((k_1 −1)/(k_2 −1)) (2) (y_1 /a_1 )−1=a_2 ∧(y_2 /a_2 )−1=a_1 ((a_1 k_(1 ) )/a_1 )−1=a_1 k_3 ∧((a_1 k_2 k_3 )/(a_1 k_3 ))−1=a_1 a_1 =((k_1 −1)/k_3 ) ∧ a_1 =k_2 −1 .....
$$\left(\mathrm{1}\right)\:\:{a}_{\mathrm{1}} \mid{y}_{\mathrm{1}} \wedge{a}_{\mathrm{2}} \mid{y}_{\mathrm{2}} \wedge{a}_{\mathrm{1}} \mid{a}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\:\frac{{y}_{\mathrm{1}} }{{a}_{\mathrm{1}} }−\mathrm{1}={a}_{\mathrm{2}} \wedge\frac{{y}_{\mathrm{2}} }{{a}_{\mathrm{2}} }−\mathrm{1}={a}_{\mathrm{1}} \\ $$$$ \\ $$$${y}_{\mathrm{2}} −{y}_{\mathrm{1}} ={a}_{\mathrm{2}} −{a}_{\mathrm{1}} >\mathrm{100} \\ $$$${y}_{\mathrm{1}} <\mathrm{2019} \\ $$$${y}_{\mathrm{1}} ={a}_{\mathrm{1}} {k}_{\mathrm{1}\:} ,{y}_{\mathrm{2}} ={a}_{\mathrm{2}} {k}_{\mathrm{2}} \:,{a}_{\mathrm{2}} ={a}_{\mathrm{1}} {k}_{\mathrm{3}} \:\left({Say}\right) \\ $$$$\:\:\:\:\left(\:{k}_{\mathrm{1}} ,{k}_{\mathrm{2}} ,{k}_{\mathrm{3}} \in\mathbb{N}\right) \\ $$$${y}_{\mathrm{1}} ={a}_{\mathrm{1}} {k}_{\mathrm{1}\:} ,{y}_{\mathrm{2}} =\left({a}_{\mathrm{1}} {k}_{\mathrm{3}} \right){k}_{\mathrm{2}} \:={a}_{\mathrm{1}} {k}_{\mathrm{2}} {k}_{\mathrm{3}} \\ $$$${y}_{\mathrm{2}} −{y}_{\mathrm{1}} ={a}_{\mathrm{2}} −{a}_{\mathrm{1}} >\mathrm{100}\:\:\left({y}_{\mathrm{2}} \:{is}\:{very}\:{high}\right) \\ $$$$\:\:\:\:\Rightarrow{a}_{\mathrm{1}} {k}_{\mathrm{2}} {k}_{\mathrm{3}} −{a}_{\mathrm{1}} {k}_{\mathrm{1}\:} ={a}_{\mathrm{1}} {k}_{\mathrm{3}} −{a}_{\mathrm{1}} \\ $$$$\:\:\:\:\Rightarrow{k}_{\mathrm{2}} {k}_{\mathrm{3}} −{k}_{\mathrm{1}\:} ={k}_{\mathrm{3}} −\mathrm{1} \\ $$$$\:\:\:\:\Rightarrow{k}_{\mathrm{3}} \left({k}_{\mathrm{2}} −\mathrm{1}\right)={k}_{\mathrm{1}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{k}_{\mathrm{3}} =\frac{{k}_{\mathrm{1}} −\mathrm{1}}{{k}_{\mathrm{2}} −\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\:\:\frac{{y}_{\mathrm{1}} }{{a}_{\mathrm{1}} }−\mathrm{1}={a}_{\mathrm{2}} \wedge\frac{{y}_{\mathrm{2}} }{{a}_{\mathrm{2}} }−\mathrm{1}={a}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\frac{{a}_{\mathrm{1}} {k}_{\mathrm{1}\:} }{{a}_{\mathrm{1}} }−\mathrm{1}={a}_{\mathrm{1}} {k}_{\mathrm{3}} \wedge\frac{{a}_{\mathrm{1}} {k}_{\mathrm{2}} {k}_{\mathrm{3}} }{{a}_{\mathrm{1}} {k}_{\mathrm{3}} }−\mathrm{1}={a}_{\mathrm{1}} \\ $$$$\:\:\:\:\:{a}_{\mathrm{1}} =\frac{{k}_{\mathrm{1}} −\mathrm{1}}{{k}_{\mathrm{3}} }\:\wedge\:{a}_{\mathrm{1}} ={k}_{\mathrm{2}} −\mathrm{1} \\ $$$$….. \\ $$
Commented by MJS last updated on 07/Nov/19
thanks for trying so we need more information...
$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{trying} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{need}\:\mathrm{more}\:\mathrm{information}… \\ $$

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