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Question Number 77603 by kaivan.ahmadi last updated on 08/Jan/20
∫((10x^2 −8x+1)/(x^4 −x^3 −x+1))dx
$$\int\frac{\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}}{dx} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 08/Jan/20
let I =∫ ((10x^2 −8x +1)/(x^4 −x^3 −x+1))dx⇒I=∫ F(x)dx we have x^4 −x^3 −x+1= x^3 (x−1)−(x−1) =(x−1)(x^3 −1)=(x−1)^2 (x^2 +x+1) ⇒ F(x)=((10x^2 −8x+1)/((x−1)^2 (x^2 +x+1))) =(a/(x−1)) +(b/((x−1)^2 )) +((cx+d)/(x^2 +x+1)) b=(x−1)^2 F(x)∣_(x=1) =(3/3)=1 lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x)=(a/(x−1))+(1/((x−1)^2 )) +((−ax+d)/(x^2 +x+1)) F(0)=1=−a +1 +d ⇒d=a ⇒F(x)=(a/(x−1))+(1/((x−1)^2 ))+((−ax+a)/(x^2 +x+1)) F(−1)=((19)/4) =−(a/2)+(1/4) +((2a)/1) ⇒19=−2a+1+8a ⇒18=6a ⇒a=3 ⇒F(x)=(3/(x−1)) +(1/((x−1)^2 )) +((−3x+3)/(x^2 +x+1)) ⇒ I =3 ∫ (dx/(x−1)) +∫ (dx/((x−1)^2 ))−(3/2) ∫ ((2x−2)/(x^2 +x+1))dx =3ln∣x−1∣−(1/(x−1))−(3/2) ∫ ((2x+1−3)/(x^2 +x+1))dx =3ln∣x−1∣−(1/(x−1))−(3/2)ln(x^2 +x+1)+(9/2) ∫ (dx/(x^2 +x+1)) we have ∫ (dx/(x^2 +x+1)) =∫ (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)))) ⇒ I =3ln∣x−1∣−(1/(x−1)) +3(√3) arctan(((2x+1)/( (√3)))) +C .
$${let}\:{I}\:=\int\:\frac{\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}}{dx}\Rightarrow{I}=\int\:{F}\left({x}\right){dx}\:{we}\:{have}\:{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}= \\ $$$${x}^{\mathrm{3}} \left({x}−\mathrm{1}\right)−\left({x}−\mathrm{1}\right)\:=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} −\mathrm{1}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)}\:=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$${b}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)\mid_{{x}=\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{3}}=\mathrm{1} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−{ax}+{d}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{1}=−{a}\:+\mathrm{1}\:+{d}\:\Rightarrow{d}={a}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{−{ax}+{a}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}} \\ $$$${F}\left(−\mathrm{1}\right)=\frac{\mathrm{19}}{\mathrm{4}}\:=−\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{2}{a}}{\mathrm{1}}\:\:\Rightarrow\mathrm{19}=−\mathrm{2}{a}+\mathrm{1}+\mathrm{8}{a}\:\Rightarrow\mathrm{18}=\mathrm{6}{a}\:\Rightarrow{a}=\mathrm{3} \\ $$$$\Rightarrow{F}\left({x}\right)=\frac{\mathrm{3}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−\mathrm{3}{x}+\mathrm{3}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=\mathrm{3}\:\int\:\frac{{dx}}{{x}−\mathrm{1}}\:+\int\:\:\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{2}}\:\int\:\frac{\mathrm{2}{x}−\mathrm{2}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx} \\ $$$$=\mathrm{3}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}}\:\int\:\frac{\mathrm{2}{x}+\mathrm{1}−\mathrm{3}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx} \\ $$$$=\mathrm{3}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)+\frac{\mathrm{9}}{\mathrm{2}}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:\:{we}\:{have} \\ $$$$\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:=\int\:\:\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \frac{\mathrm{4}}{\mathrm{3}}\:\:\int\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{3}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\mathrm{3}\sqrt{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+{C}\:. \\ $$
Answered by Kunal12588 last updated on 08/Jan/20
x^4 −x^3 −x+1=x(x^3 −1)−(x^3 −1) =(x−1)(x^3 −1) =(x−1)^2 (x^2 +x+1) ((10x^2 −8x+1)/((x−1)^2 (x^2 +x+1)))=(a/((x−1)))+(b/((x−1)^2 ))+((cx+d)/((x^2 +x+1))) ⇒10x^2 −8x+1=a(x^4 −x^3 −x+1)+b(x^3 −1) +(cx+d)(x−1)^3 10x^2 −8x+1= { ((ax^4 +(−a+b)x^3 −ax+(a−b))),((+(cx+d)(x^3 −3x^2 +3x−1))) :} 10x^2 −8x+1= { ((ax^4 +(−a+b)x^3 −ax+(a−b))),((+cx^4 +(−3c+d)x^3 +(3c−3d)x^2 )),((+(−c+3d)x−d)) :} 10x^2 −8x+1= { (((a+c)x^4 +(−a+b−3c+d)x^3 )),((+(3c−3d)x^2 +(−a−c+3d)x)),((+(a−b−d))) :} 1) a+c=0 2) −a+b−3c+d=0 3) 3c−3d=10 4) −a−c+3d=−8 5) a−b−d=1 1⇒c=−a ⇒ { ((b+d=−2a)),((−3a−3d=10⇒d=−((3a+10)/3))),((d=((−8)/3))) :} ⇒ { ((b=−2a+(8/3))),((3a+10=8⇒a=6⇒c=−6)) :} ⇒b=−12+(8/3)=((−36+8)/3)=((−28)/3) ((10x^2 −8x+1)/((x−1)^2 (x^2 +x+1)))=(6/((x−1)))−((28)/(3(x−1)^2 ))−((18x+8)/(3(x^2 +x+1))) ∫((10x^2 −8x+1)/(x^4 −x^3 −x+1))dx =6∫(dx/(x−1))−((28)/3)∫(dx/((x−1)^2 ))−(1/3)∫((18x+9−1)/(x^2 +x+1))dx =6log∣x−1∣+((28)/(3(x−1)))−(9/3)log∣x^2 +x+1∣+(1/3)∫(dx/((x+(1/2))^2 +(3/4))) =6log∣x−1∣+((28)/(3(x−1)))−3log∣x^2 +x+1∣+(1/3)∫(dx/((x+(1/2))^2 +(((√3)/2))^2 )) =6log∣x−1∣+((28)/(3(x−1)))−3log∣x^2 +x+1∣+(1/3)×(2/( (√3))) tan^(−1) (((x+(1/2))/((√3)/2)))+C =6log∣x−1∣+((28)/(3(x−1)))−3log∣x^2 +x+1∣+(2/(3(√3))) tan^(−1) (((2x+1)/( (√3))))+C
$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}={x}\left({x}^{\mathrm{3}} −\mathrm{1}\right)−\left({x}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}=\frac{{a}}{\left({x}−\mathrm{1}\right)}+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{cx}+{d}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}={a}\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}\right)+{b}\left({x}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({cx}+{d}\right)\left({x}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}=\begin{cases}{{ax}^{\mathrm{4}} +\left(−{a}+{b}\right){x}^{\mathrm{3}} −{ax}+\left({a}−{b}\right)}\\{+\left({cx}+{d}\right)\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}\right)}\end{cases} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}=\begin{cases}{{ax}^{\mathrm{4}} +\left(−{a}+{b}\right){x}^{\mathrm{3}} −{ax}+\left({a}−{b}\right)}\\{+{cx}^{\mathrm{4}} +\left(−\mathrm{3}{c}+{d}\right){x}^{\mathrm{3}} +\left(\mathrm{3}{c}−\mathrm{3}{d}\right){x}^{\mathrm{2}} }\\{+\left(−{c}+\mathrm{3}{d}\right){x}−{d}}\end{cases} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}=\begin{cases}{\left({a}+{c}\right){x}^{\mathrm{4}} +\left(−{a}+{b}−\mathrm{3}{c}+{d}\right){x}^{\mathrm{3}} }\\{+\left(\mathrm{3}{c}−\mathrm{3}{d}\right){x}^{\mathrm{2}} +\left(−{a}−{c}+\mathrm{3}{d}\right){x}}\\{+\left({a}−{b}−{d}\right)}\end{cases} \\ $$$$\left.\mathrm{1}\right)\:\:{a}+{c}=\mathrm{0}\: \\ $$$$\left.\mathrm{2}\right)\:\:−{a}+{b}−\mathrm{3}{c}+{d}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\:\mathrm{3}{c}−\mathrm{3}{d}=\mathrm{10} \\ $$$$\left.\mathrm{4}\right)\:\:−{a}−{c}+\mathrm{3}{d}=−\mathrm{8} \\ $$$$\left.\mathrm{5}\right)\:{a}−{b}−{d}=\mathrm{1} \\ $$$$\mathrm{1}\Rightarrow{c}=−{a} \\ $$$$\Rightarrow\begin{cases}{{b}+{d}=−\mathrm{2}{a}}\\{−\mathrm{3}{a}−\mathrm{3}{d}=\mathrm{10}\Rightarrow{d}=−\frac{\mathrm{3}{a}+\mathrm{10}}{\mathrm{3}}}\\{{d}=\frac{−\mathrm{8}}{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{b}=−\mathrm{2}{a}+\frac{\mathrm{8}}{\mathrm{3}}}\\{\mathrm{3}{a}+\mathrm{10}=\mathrm{8}\Rightarrow{a}=\mathrm{6}\Rightarrow{c}=−\mathrm{6}}\end{cases} \\ $$$$\Rightarrow{b}=−\mathrm{12}+\frac{\mathrm{8}}{\mathrm{3}}=\frac{−\mathrm{36}+\mathrm{8}}{\mathrm{3}}=\frac{−\mathrm{28}}{\mathrm{3}} \\ $$$$\frac{\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}=\frac{\mathrm{6}}{\left({x}−\mathrm{1}\right)}−\frac{\mathrm{28}}{\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{18}{x}+\mathrm{8}}{\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$$\int\frac{\mathrm{10}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}}{dx} \\ $$$$=\mathrm{6}\int\frac{{dx}}{{x}−\mathrm{1}}−\frac{\mathrm{28}}{\mathrm{3}}\int\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{18}{x}+\mathrm{9}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$=\mathrm{6}{log}\mid{x}−\mathrm{1}\mid+\frac{\mathrm{28}}{\mathrm{3}\left({x}−\mathrm{1}\right)}−\frac{\mathrm{9}}{\mathrm{3}}{log}\mid{x}^{\mathrm{2}} +{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=\mathrm{6}{log}\mid{x}−\mathrm{1}\mid+\frac{\mathrm{28}}{\mathrm{3}\left({x}−\mathrm{1}\right)}−\mathrm{3}{log}\mid{x}^{\mathrm{2}} +{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{6}{log}\mid{x}−\mathrm{1}\mid+\frac{\mathrm{28}}{\mathrm{3}\left({x}−\mathrm{1}\right)}−\mathrm{3}{log}\mid{x}^{\mathrm{2}} +{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)+{C} \\ $$$$=\mathrm{6}{log}\mid{x}−\mathrm{1}\mid+\frac{\mathrm{28}}{\mathrm{3}\left({x}−\mathrm{1}\right)}−\mathrm{3}{log}\mid{x}^{\mathrm{2}} +{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$
Commented by Kunal12588 last updated on 08/Jan/20
next time try yourself first !
$${next}\:{time}\:{try}\:{yourself}\:{first}\:! \\ $$
Commented by key of knowledge last updated on 08/Jan/20
log is not,ln is true (∫(du/u)=lnx)
$$\mathrm{log}\:\mathrm{is}\:\mathrm{not},\mathrm{ln}\:\mathrm{is}\:\mathrm{true}\:\:\left(\int\frac{\mathrm{du}}{\mathrm{u}}=\mathrm{lnx}\right) \\ $$
Commented by kaivan.ahmadi last updated on 08/Jan/20
thank u sir
$${thank}\:{u}\:{sir} \\ $$
Commented by MJS last updated on 08/Jan/20
some write log for ln, and log_(10) others write log for log_(10) and ln in case of integrals there had been no case yet where log meant log_(10)
$$\mathrm{some}\:\mathrm{write}\:\mathrm{log}\:\mathrm{for}\:\mathrm{ln},\:\mathrm{and}\:\mathrm{log}_{\mathrm{10}} \\ $$$$\mathrm{others}\:\mathrm{write}\:\mathrm{log}\:\mathrm{for}\:\mathrm{log}_{\mathrm{10}} \:\mathrm{and}\:\mathrm{ln} \\ $$$$\mathrm{in}\:\mathrm{case}\:\mathrm{of}\:\mathrm{integrals}\:\mathrm{there}\:\mathrm{had}\:\mathrm{been}\:\mathrm{no}\:\mathrm{case} \\ $$$$\mathrm{yet}\:\mathrm{where}\:\mathrm{log}\:\mathrm{meant}\:\mathrm{log}_{\mathrm{10}} \\ $$
Commented by mathmax by abdo last updated on 08/Jan/20
by notation Log means ln and log means log_(10) ...
$${by}\:{notation}\:{Log}\:{means}\:{ln}\:{and}\:{log}\:{means}\:{log}_{\mathrm{10}} \:… \\ $$
Commented by Kunal12588 last updated on 09/Jan/20
yeah, there is much of confusion but I am just following what I learned in my school and NCERT book. log a → log_e a and for referring to log_(10) a we write it like that log_(10) a. nothing mentioned about ln.
$${yeah},\:{there}\:{is}\:{much}\:{of}\:{confusion} \\ $$$${but}\:{I}\:{am}\:{just}\:{following}\:{what}\:{I}\:{learned}\:{in} \\ $$$${my}\:{school}\:{and}\:{NCERT}\:\:{book}. \\ $$$${log}\:{a}\:\rightarrow\:\:{log}_{{e}} {a}\:\:{and}\:{for}\:{referring}\:{to}\:{log}_{\mathrm{10}} {a} \\ $$$${we}\:{write}\:{it}\:{like}\:{that}\:{log}_{\mathrm{10}} {a}. \\ $$$${nothing}\:{mentioned}\:{about}\:{ln}. \\ $$

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