Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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...

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{trying} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{need}\:\mathrm{more}\:\mathrm{information}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com