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Question Number 67035 by rajesh4661kumar@gmail.com last updated on 22/Aug/19

Commented by rajesh4661kumar@gmail.com last updated on 22/Aug/19

$s o l v e p l e a s e a r g e n t h a i$
Commented by mathmax by abdo last updated on 22/Aug/19
$let I =∫ (dx/(x^4 −x^2 +1)) ⇒I =∫ (dx/((x^2 +1)^2 −3x^2 ))=∫ (dx/((x^2 +1−(√3)x)(x^2 +1+(√3)x))) let decompose F(x)=(1/((x^2 −(√3)x +1)(x^2 +(√3)x +1))) ⇒ F(x)=((ax+b)/(x^2 −(√3)x +1)) +((cx+d)/(x^2 +(√3)x +1)) F(−x)=F(x) ⇒ ((−ax+b)/(x^2 +(√3)x+1)) +((−cx+d)/(x^2 −(√3)x +1)) =F(x) ⇒c=−a and b=d ⇒F(x) =((ax+b)/(x^2 −(√3)x +1)) +((−ax +b)/(x^2 +(√3)x +1)) F(0) =1 =2b ⇒b=(1/2) F(1) =((a+b)/(2−(√3))) +((−a+b)/(2+(√3))) =1 ⇒(2+(√3))(a+(1/2))+(2−(√3))(−a+(1/2))=1⇒ ⇒(2+(√3))a +((2+(√3))/2) −(2−(√3))a +((2−(√3))/2) =1 ⇒ 2(√3)a +2 =1 ⇒2(√3)a =−1 ⇒ a =−(1/(2(√3))) ⇒ F(x) =(1/2) ((−(1/(√3))x+1)/(x^2 −(√3)x+1)) +(1/2)(((1/((√3) ))x+1)/(x^2 +(√3)x +1)) ⇒ ∫ F(x)dx =(1/(2(√3))) ∫ ((x+(√3))/(x^2 +(√3)x +1))−(1/(2(√3))) ∫ ((x−(√3))/(x^2 −(√3)x+1)) =H−K H=∫ ((x+(√3))/(x^2 +(√3)x +1)) =(1/2) ∫ ((2x+2(√3))/(x^2 +(√3)x +1)) =(1/2)ln(x^2 +(√3)x +1)+((√3)/2)∫(dx/(x^2 +(√3)x +1)) ∫ (dx/(x^2 +(√3)x +1)) =∫ (dx/(x^2 +2((√3)/2)x +(3/4) +1−(3/4))) =∫ (dx/((x+((√3)/2))^2 +(1/4))) =_(x+((√3)/2)=(t/2)) 4∫ (1/(t^2 +1)) (dt/2) =2 arctan(2x+(√3)) ⇒ H =(1/(2(√3)))×(1/2)ln(x^2 +(√3)x +1)+(1/(2(√3)))×((√3)/2) ×2 arctan(2x+(√3)) =(1/(4(√3)))ln(x^2 +(√3)x +1)+(1/2) arctan(2x+(√3)) K=(1/(2(√3)))∫ ((x−(√3))/(x^2 −(√3)x+1)) dx =_(x =−t) (1/(2(√3))) ∫ ((−t−(√3))/(t^2 +(√3)t +1))(−dt) =(1/(2(√3)))∫ ((t+(√3))/(t^2 +(√3)t +1))dt =(1/(4(√3)))ln(x^2 −(√3)x+1)+(1/2)arctan(−2x+(√3)) ⇒ I =(1/(4(√3)))ln(x^2 +(√3)x+1)+(1/2) arctan(2x+(√3)) −(1/(4(√3)))ln(x^2 −(√3)x+1)+(1/2) arctan(2x−(√3)) +C I=(1/(4(√3)))ln(((x^2 +(√3)x +1)/(x^2 −(√3)x +1))) +(1/2){ arctan(2x+(√3)) +arctan(2x−(√3))} +C$
$l e t I = \int \frac{d x}{x^{4} - x^{2} + 1} \Rightarrow I = \int \frac{d x}{{(x^{2} + 1)}^{2} - 3 x^{2}} = \int \frac{d x}{(x^{2} + 1 - \sqrt{3} x) (x^{2} + 1 + \sqrt{3} x)}$ $l e t d e c o m p o s e F (x) = \frac{1}{(x^{2} - \sqrt{3} x + 1) (x^{2} + \sqrt{3} x + 1)} \Rightarrow$ $F (x) = \frac{a x + b}{x^{2} - \sqrt{3} x + 1} + \frac{c x + d}{x^{2} + \sqrt{3} x + 1}$ $F (- x) = F (x) \Rightarrow \frac{- a x + b}{x^{2} + \sqrt{3} x + 1} + \frac{- c x + d}{x^{2} - \sqrt{3} x + 1} = F (x) \Rightarrow c = - a a n d b = d$ $\Rightarrow F (x) = \frac{a x + b}{x^{2} - \sqrt{3} x + 1} + \frac{- a x + b}{x^{2} + \sqrt{3} x + 1}$ $F (0) = 1 = 2 b \Rightarrow b = \frac{1}{2}$ $F (1) = \frac{a + b}{2 - \sqrt{3}} + \frac{- a + b}{2 + \sqrt{3}} = 1 \Rightarrow (2 + \sqrt{3}) (a + \frac{1}{2}) + (2 - \sqrt{3}) (- a + \frac{1}{2}) = 1 \Rightarrow$ $\Rightarrow (2 + \sqrt{3}) a + \frac{2 + \sqrt{3}}{2} - (2 - \sqrt{3}) a + \frac{2 - \sqrt{3}}{2} = 1 \Rightarrow$ $2 \sqrt{3} a + 2 = 1 \Rightarrow 2 \sqrt{3} a = - 1 \Rightarrow a = - \frac{1}{2 \sqrt{3}} \Rightarrow$ $F (x) = \frac{1}{2} \frac{- \frac{1}{\sqrt{3}} x + 1}{x^{2} - \sqrt{3} x + 1} + \frac{1}{2} \frac{\frac{1}{\sqrt{3}} x + 1}{x^{2} + \sqrt{3} x + 1} \Rightarrow$ $\int F (x) d x = \frac{1}{2 \sqrt{3}} \int \frac{x + \sqrt{3}}{x^{2} + \sqrt{3} x + 1} - \frac{1}{2 \sqrt{3}} \int \frac{x - \sqrt{3}}{x^{2} - \sqrt{3} x + 1} = H - K$ $H = \int \frac{x + \sqrt{3}}{x^{2} + \sqrt{3} x + 1} = \frac{1}{2} \int \frac{2 x + 2 \sqrt{3}}{x^{2} + \sqrt{3} x + 1} = \frac{1}{2} l n (x^{2} + \sqrt{3} x + 1) + \frac{\sqrt{3}}{2} \int \frac{d x}{x^{2} + \sqrt{3} x + 1}$ $\int \frac{d x}{x^{2} + \sqrt{3} x + 1} = \int \frac{d x}{x^{2} + 2 \frac{\sqrt{3}}{2} x + \frac{3}{4} + 1 - \frac{3}{4}} = \int \frac{d x}{{(x + \frac{\sqrt{3}}{2})}^{2} + \frac{1}{4}}$ $=_{x + \frac{\sqrt{3}}{2} = \frac{t}{2}} 4 \int \frac{1}{t^{2} + 1} \frac{d t}{2} = 2 a r c t a n (2 x + \sqrt{3}) \Rightarrow$ $H = \frac{1}{2 \sqrt{3}} \times \frac{1}{2} l n (x^{2} + \sqrt{3} x + 1) + \frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{2} \times 2 a r c t a n (2 x + \sqrt{3})$ $= \frac{1}{4 \sqrt{3}} l n (x^{2} + \sqrt{3} x + 1) + \frac{1}{2} a r c t a n (2 x + \sqrt{3})$ $K = \frac{1}{2 \sqrt{3}} \int \frac{x - \sqrt{3}}{x^{2} - \sqrt{3} x + 1} d x =_{x = - t} \frac{1}{2 \sqrt{3}} \int \frac{- t - \sqrt{3}}{t^{2} + \sqrt{3} t + 1} (- d t) = \frac{1}{2 \sqrt{3}} \int \frac{t + \sqrt{3}}{t^{2} + \sqrt{3} t + 1} d t$ $= \frac{1}{4 \sqrt{3}} l n (x^{2} - \sqrt{3} x + 1) + \frac{1}{2} a r c t a n (- 2 x + \sqrt{3}) \Rightarrow$ $I = \frac{1}{4 \sqrt{3}} l n (x^{2} + \sqrt{3} x + 1) + \frac{1}{2} a r c t a n (2 x + \sqrt{3})$ $- \frac{1}{4 \sqrt{3}} l n (x^{2} - \sqrt{3} x + 1) + \frac{1}{2} a r c t a n (2 x - \sqrt{3}) + C$ $I = \frac{1}{4 \sqrt{3}} l n (\frac{x^{2} + \sqrt{3} x + 1}{x^{2} - \sqrt{3} x + 1}) + \frac{1}{2} {a r c t a n (2 x + \sqrt{3}) + a r c t a n (2 x - \sqrt{3})} + C$
	Answered by Tanmay chaudhury last updated on 22/Aug/19

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