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Question Number 132333 by liberty last updated on 13/Feb/21
very nice integral ∫ ((4x^3 +4x^2 +4x+3)/((x^2 +1)(x^2 +x+1)^2 )) dx?
$$\:\mathrm{very}\:\mathrm{nice}\:\mathrm{integral} \\ $$$$\int\:\frac{\mathrm{4x}^{\mathrm{3}} +\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{3}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}? \\ $$$$ \\ $$
Answered by EDWIN88 last updated on 13/Feb/21
well... let me try Ostrogradsky method the integral has form ∫ ((4x^3 +4x^2 +4x+3)/((x^2 +1)(x^2 +x+1)^2 )) dx=((ax+b)/(x^2 +x+1)) +∫ ((cx^3 +dx^2 +ex+f)/((x^2 +1)(x^2 +x+1))) dx differentiating both sides ((4x^3 +4x^2 +4x+3)/((x^2 +1)(x^2 +x+1)^2 )) = ((a(x^2 +x+1)−(ax+b)(2x+1))/((x^2 +x+1)^2 ))+((cx^3 +dx^2 +ex+f)/((x^2 +1)(x^2 +x+1))) after solving the coefficients we get a=1 , b=−1 ,c=0 , d=1 ,e=1 ,f = 1 then I = ((x−1)/(x^2 +x+1)) +∫ ((x^2 +x+1)/((x^2 +1)(x^2 +x+1)))dx I= ((x−1)/(x^2 +x+1)) + ∫ (dx/(x^2 +1))= ((x−1)/(x^2 +x+1))+ arctan (x) + C
$$\mathrm{well}…\:\mathrm{let}\:\mathrm{me}\:\mathrm{try} \\ $$$$\mathrm{Ostrogradsky}\:\mathrm{method} \\ $$$$\mathrm{the}\:\mathrm{integral}\:\mathrm{has}\:\mathrm{form}\: \\ $$$$\int\:\frac{\mathrm{4x}^{\mathrm{3}} +\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{3}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:+\int\:\frac{\mathrm{cx}^{\mathrm{3}} +\mathrm{dx}^{\mathrm{2}} +\mathrm{ex}+\mathrm{f}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)}\:\mathrm{dx} \\ $$$$\mathrm{differentiating}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\:\frac{\mathrm{4x}^{\mathrm{3}} +\mathrm{4x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{3}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{a}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)−\left(\mathrm{ax}+\mathrm{b}\right)\left(\mathrm{2x}+\mathrm{1}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{cx}^{\mathrm{3}} +\mathrm{dx}^{\mathrm{2}} +\mathrm{ex}+\mathrm{f}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)} \\ $$$$\mathrm{after}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{coefficients}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{a}=\mathrm{1}\:,\:\mathrm{b}=−\mathrm{1}\:,\mathrm{c}=\mathrm{0}\:,\:\mathrm{d}=\mathrm{1}\:,\mathrm{e}=\mathrm{1}\:,\mathrm{f}\:=\:\mathrm{1} \\ $$$$\mathrm{then}\:\mathrm{I}\:=\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:+\int\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$$$\mathrm{I}=\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:+\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}=\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}+\:\mathrm{arctan}\:\left(\mathrm{x}\right)\:+\:\mathrm{C} \\ $$
Commented by liberty last updated on 13/Feb/21
scho^ ne Lo^ sung
$$ \\ $$$$\mathrm{sch}\ddot {\mathrm{o}ne}\:\mathrm{L}\ddot {\mathrm{o}sung} \\ $$
Commented by SLVR last updated on 13/Feb/21
great job...nice sir
$${great}\:{job}…{nice}\:{sir} \\ $$
Commented by SLVR last updated on 13/Feb/21
but on differentiaion of RHS 1st part denominator has no sqare...how...to proceed..
$${but}\:{on}\:{differentiaion}\:{of}\:{RHS}\:\mathrm{1}{st}\:{part}\:{denominator} \\ $$$${has}\:{no}\:{sqare}…{how}…{to}\:{proceed}.. \\ $$
Commented by EDWIN88 last updated on 13/Feb/21
oh it is only typo
$$\mathrm{oh}\:\mathrm{it}\:\mathrm{is}\:\mathrm{only}\:\mathrm{typo} \\ $$
Commented by SLVR last updated on 13/Feb/21
welcome sir..
$${welcome}\:{sir}.. \\ $$
Answered by EDWIN88 last updated on 13/Feb/21
∫ ((P(x))/(Q(x))) dx = ((P_1 (x))/(Q_1 (x))) + ∫ ((P_2 (x))/(Q_2 (x))) dx where Q_1 (x)is gcd of Q(x) and Q′(x) Q_2 (x)=((Q(x))/(Q_1 (x))) and P_1 (x) and P_2 (x) are unknown polynomials of degree one less than Q_1 (x) and Q_2 (x) respectively . from the equation Q(x)=(x^2 +1)(x^2 +x+1)^2 Q′(x)=2x(x^2 +x+1)^2 +2(x^2 +1)(2x+1)(x^2 +x+1) the gcd of Q(x) and Q′(x) is Q_1 (x)=x^2 +x+1 and Q_2 (x)=((Q(x))/(Q_1 (x)))= (x^2 +1)(x^2 +x+1)
$$\int\:\frac{\mathrm{P}\left(\mathrm{x}\right)}{\mathrm{Q}\left(\mathrm{x}\right)}\:\mathrm{dx}\:=\:\frac{\mathrm{P}_{\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{Q}_{\mathrm{1}} \left(\mathrm{x}\right)}\:+\:\int\:\frac{\mathrm{P}_{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{Q}_{\mathrm{2}} \left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$$$\mathrm{where}\:\mathrm{Q}_{\mathrm{1}} \left(\mathrm{x}\right)\mathrm{is}\:\mathrm{gcd}\:\mathrm{of}\:\mathrm{Q}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{Q}'\left(\mathrm{x}\right) \\ $$$$\mathrm{Q}_{\mathrm{2}} \left(\mathrm{x}\right)=\frac{\mathrm{Q}\left(\mathrm{x}\right)}{\mathrm{Q}_{\mathrm{1}} \left(\mathrm{x}\right)}\:\mathrm{and}\:\mathrm{P}_{\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{P}_{\mathrm{2}} \left(\mathrm{x}\right)\:\mathrm{are}\: \\ $$$$\mathrm{unknown}\:\mathrm{polynomials}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{one}\:\mathrm{less} \\ $$$$\mathrm{than}\:\mathrm{Q}_{\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{Q}_{\mathrm{2}} \left(\mathrm{x}\right)\:\mathrm{respectively}\:.\: \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{Q}\left(\mathrm{x}\right)=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{Q}'\left(\mathrm{x}\right)=\mathrm{2x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{the}\:\mathrm{gcd}\:\mathrm{of}\:\mathrm{Q}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{Q}'\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{Q}_{\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{Q}_{\mathrm{2}} \left(\mathrm{x}\right)=\frac{\mathrm{Q}\left(\mathrm{x}\right)}{\mathrm{Q}_{\mathrm{1}} \left(\mathrm{x}\right)}=\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right) \\ $$

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