∫ ((4x^2 + 3)/((x^2 + x + 1)^2 )) dx


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Question Number 52649 by Tawa1 last updated on 10/Jan/19

$\int \frac{{4 x}^{2} + 3}{{(x^{2} + x + 1)}^{2}} dx$
Commented by maxmathsup by imad last updated on 10/Jan/19
$et I =∫ ((4x^2 +3)/((x^2 +x+1)^2 ))dx ⇒ I =∫ ((4(x^2 +x+1)−4x−1)/((x^2 +x+1)^2 ))dx =4 ∫ (dx/(x^2 +x +1)) −∫ ((4x+1)/((x^2 +x+1)^2 ))dx =4H −K H =∫ (dx/((x+(1/2))^2 +(3/4))) =_(x+(1/2)=((√3)/2)tanθ) ∫ (1/((3/4)(1+tan^2 θ))) ((√3)/2) (1+tan^2 θ)dθ =((√3)/2) (4/3) ∫ dθ =(2/(√3)) θ +c_1 =(2/(√3)) arctan(((2x+1)/(√3))) +c_1 we have ((1/(x^2 +x+1)))^′ =−((2x+1)/((x^2 +x +1))) ⇒ ∫ ((4x+1)/((x^2 +x+1)^2 )) =2 ∫ ((2x+(1/2))/((x^2 +x+1)^2 ))dx =2 ∫ ((2x+1−(1/2))/((x^2 +x+1)^2 ))dx =2{−(1/(x^2 +x+1))−(1/2) ∫ (dx/((x^2 +x+1)^2 ))} =((−2)/(x^2 +x+1)) −∫ (dx/((x^2 +x+1)^2 )) but ∫ (dx/((x^2 +x+1)^2 )) =∫ (dx/(((x+(1/2))^2 +(3/4))^2 )) =_(x+(1/2)=((√3)/2)tanθ) ((16)/9)((√3)/2) ∫ (1/((1+tan^2 θ)^2 )) (1+tan^2 θ)dθ =((8(√3))/9) ∫ (dθ/(1+tan^2 θ)) =((8(√3))/9) ∫ ((1+cos(2θ))/2)dθ =((4(√3))/9) ∫ (1+cos(2θ))dθ =((4(√3))/9) θ +((4(√3))/(18)) sin(2θ) =((4(√3))/9) arctan(((2x+1)/(√3))) +((2(√3))/9) sin(2arctan(((2x+1)/(√3))))+c_2 ⇒ I =(8/(√3)) arctan(((2x+1)/(√3))) +(2/(x^2 +x+1)) +((4(√3))/9) arctan(((2x+1)/(√3)))+((2(√3))/9) sin(2arctan(((2x+1)/(√3)))) +C$
$e t I = \int \frac{4 x^{2} + 3}{{(x^{2} + x + 1)}^{2}} d x \Rightarrow I = \int \frac{4 (x^{2} + x + 1) - 4 x - 1}{{(x^{2} + x + 1)}^{2}} d x$ $= 4 \int \frac{d x}{x^{2} + x + 1} - \int \frac{4 x + 1}{{(x^{2} + x + 1)}^{2}} d x = 4 H - K$ $H = \int \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n θ} \int \frac{1}{\frac{3}{4} (1 + {t a n}^{2} θ)} \frac{\sqrt{3}}{2} (1 + {t a n}^{2} θ) d θ$ $= \frac{\sqrt{3}}{2} \frac{4}{3} \int d θ = \frac{2}{\sqrt{3}} θ + c_{1} = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + c_{1}$ $w e h a v e {(\frac{1}{x^{2} + x + 1})}^{^{'}} = - \frac{2 x + 1}{(x^{2} + x + 1)} \Rightarrow$ $\int \frac{4 x + 1}{{(x^{2} + x + 1)}^{2}} = 2 \int \frac{2 x + \frac{1}{2}}{{(x^{2} + x + 1)}^{2}} d x = 2 \int \frac{2 x + 1 - \frac{1}{2}}{{(x^{2} + x + 1)}^{2}} d x$ $= 2 {- \frac{1}{x^{2} + x + 1} - \frac{1}{2} \int \frac{d x}{{(x^{2} + x + 1)}^{2}}}$ $= \frac{- 2}{x^{2} + x + 1} - \int \frac{d x}{{(x^{2} + x + 1)}^{2}} b u t$ $\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = \int \frac{d x}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{2}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n θ} \frac{16}{9} \frac{\sqrt{3}}{2} \int \frac{1}{{(1 + {t a n}^{2} θ)}^{2}} (1 + {t a n}^{2} θ) d θ$ $= \frac{8 \sqrt{3}}{9} \int \frac{d θ}{1 + {t a n}^{2} θ} = \frac{8 \sqrt{3}}{9} \int \frac{1 + c o s (2 θ)}{2} d θ = \frac{4 \sqrt{3}}{9} \int (1 + c o s (2 θ)) d θ$ $= \frac{4 \sqrt{3}}{9} θ + \frac{4 \sqrt{3}}{18} s i n (2 θ) = \frac{4 \sqrt{3}}{9} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{2 \sqrt{3}}{9} s i n (2 a r c t a n (\frac{2 x + 1}{\sqrt{3}})) + c_{2} \Rightarrow$ $I = \frac{8}{\sqrt{3}} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{2}{x^{2} + x + 1} + \frac{4 \sqrt{3}}{9} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{2 \sqrt{3}}{9} s i n (2 a r c t a n (\frac{2 x + 1}{\sqrt{3}})) + C$
Commented by Tawa1 last updated on 10/Jan/19

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