1)find ∫ (dx/((x^2 +x+1)^6 )) 2)calculate ∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^6 ))


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Question Number 82971 by mathmax by abdo last updated on 26/Feb/20

$1) f i n d \int \frac{d x}{{(x^{2} + x + 1)}^{6}}$ $2) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{6}}$
Commented by mathmax by abdo last updated on 26/Feb/20
$2) let A =∫_(−∞) ^(+∞) (dx/((x^2 +x+1)^6 )) ⇒ A =∫_(−∞) ^(+∞) (dx/(((x+(1/2))^2 +(3/4))^6 )) =_(x+(1/2)=((√3)/2)t) ((4/3))^6 ∫_(−∞) ^(+∞) (dt/((t^2 +1)^6 )) let also changement t=tanθ give ∫_(−∞) ^(+∞) (dt/((t^2 +1)^6 )) =∫_(−(π/2)) ^(π/2) (((1+tan^2 θ)dθ)/((1+tan^2 θ)^6 )) =∫_(−(π/2)) ^(π/2) (dθ/((1+tan^2 θ)^4 )) =∫_(−(π/2)) ^(π/2) cos^8 t dt =2 ∫_0 ^(π/2) (((1+cos(2t))/2))^4 dt =(1/8)∫_0 ^(π/2) (1+2cos(2t) +cos^2 (2t))^2 dt =(1/8)∫_0 ^(π/2) { (1+2cos(2t))^2 +2(1+2cos(2t)cos^2 (2t) +cos^4 (2t))dt =(1/8)∫_0 ^(π/2) {1+4cos(2t) +4cos^2 (2t) +2cos^2 (2t)+4cos^3 (2t)+cos^4 (2t)}dt =(π/(16)) +(1/2)∫_0 ^(π/2) cos(2t)+(3/4)∫_0 ^(π/2) cos^2 (2t)dt+(1/2)∫_0 ^(π/2) cos^3 (2t)dt+(1/8)∫_0 ^(π/2) cos^4 (2t)dt ∫_0 ^(π/2) cos(2t)dt =(1/2)sin(2t)]_0 ^(π/2) =0 ∫_0 ^(π/2) cos^2 (2t)dt =(1/2)∫_0 ^(π/2) (1+cos(4t))dt =(π/4) +(1/8)[sin(4t)]_0 ^(π/2) =(π/4) ∫_0 ^(π/2) cos^3 (2t)dt =∫_0 ^(π/2) (((1+cos(4t))/2))cos(2t)dt =(1/2)∫_0 ^(π/2) {cos(2t)+cos(2t)cos(4t)}dt =(1/2)∫_0 ^(π/2) cos(2t)dt +(1/4)∫_0 ^(π/2) (cos(2t)+cos(6t))dt=0 ∫_0 ^(π/2) cos^4 t dt =∫_0 ^(π/2) (((1+cos(2t))/2))^2 dt =(1/4)∫_0 ^(π/2) (1+2cos(2t) +cos^2 (2t))dt =(π/8) +(1/2)∫_0 ^(π/2) cos(2t)dt +(1/8)∫_0 ^(π/2) (1+cos(4t))dt =(π/8) +0+(π/(16)) =((3π)/(16)) the value of I is known$
$2) l e t A = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{6}} \Rightarrow A = \int_{- \infty}^{+ \infty} \frac{d x}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{6}}$ $=_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} {(\frac{4}{3})}^{6} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{6}} l e t a l s o c h a n g e m e n t$ $t = t a n θ g i v e \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{6}} = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} θ) d θ}{{(1 + {t a n}^{2} θ)}^{6}}$ $= \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d θ}{{(1 + {t a n}^{2} θ)}^{4}} = \int_{- \frac{π}{2}}^{\frac{π}{2}} {c o s}^{8} t d t = 2 \int_{0}^{\frac{π}{2}} {(\frac{1 + c o s (2 t)}{2})}^{4} d t$ $= \frac{1}{8} \int_{0}^{\frac{π}{2}} {(1 + 2 c o s (2 t) + {c o s}^{2} (2 t))}^{2} d t$ $= \frac{1}{8} \int_{0}^{\frac{π}{2}} {{(1 + 2 c o s (2 t))}^{2} + 2 (1 + 2 c o s (2 t) {c o s}^{2} (2 t) + {c o s}^{4} (2 t)) d t$ $= \frac{1}{8} \int_{0}^{\frac{π}{2}} {1 + 4 c o s (2 t) + 4 {c o s}^{2} (2 t) + 2 {c o s}^{2} (2 t) + 4 {c o s}^{3} (2 t) + {c o s}^{4} (2 t)} d t$ $= \frac{π}{16} + \frac{1}{2} \int_{0}^{\frac{π}{2}} c o s (2 t) + \frac{3}{4} \int_{0}^{\frac{π}{2}} {c o s}^{2} (2 t) d t + \frac{1}{2} \int_{0}^{\frac{π}{2}} {c o s}^{3} (2 t) d t + \frac{1}{8} \int_{0}^{\frac{π}{2}} {c o s}^{4} (2 t) d t$ ${\int_{0}^{\frac{π}{2}} c o s (2 t) d t = \frac{1}{2} s i n (2 t)]}_{0}^{\frac{π}{2}} = 0$ $\int_{0}^{\frac{π}{2}} {c o s}^{2} (2 t) d t = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + c o s (4 t)) d t$ $= \frac{π}{4} + \frac{1}{8} {[s i n (4 t)]}_{0}^{\frac{π}{2}} = \frac{π}{4}$ $\int_{0}^{\frac{π}{2}} {c o s}^{3} (2 t) d t = \int_{0}^{\frac{π}{2}} (\frac{1 + c o s (4 t)}{2}) c o s (2 t) d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} {c o s (2 t) + c o s (2 t) c o s (4 t)} d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} c o s (2 t) d t + \frac{1}{4} \int_{0}^{\frac{π}{2}} (c o s (2 t) + c o s (6 t)) d t = 0$ $\int_{0}^{\frac{π}{2}} {c o s}^{4} t d t = \int_{0}^{\frac{π}{2}} {(\frac{1 + c o s (2 t)}{2})}^{2} d t$ $= \frac{1}{4} \int_{0}^{\frac{π}{2}} (1 + 2 c o s (2 t) + {c o s}^{2} (2 t)) d t$ $= \frac{π}{8} + \frac{1}{2} \int_{0}^{\frac{π}{2}} c o s (2 t) d t + \frac{1}{8} \int_{0}^{\frac{π}{2}} (1 + c o s (4 t)) d t$ $= \frac{π}{8} + 0 + \frac{π}{16} = \frac{3 π}{16} t h e v a l u e o f I i s k n o w n$
Commented by abdomathmax last updated on 26/Feb/20

$1) c o m p l e x m e t h o d l e t A = \int \frac{d x}{{(x^{2} + x + 1)}^{6}}$ $x^{2} + x + 1 = 0 \to Δ = 1 - 4 = - 3 \Rightarrow$ $z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} a n d z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}$ $A = \int \frac{d x}{{(x - z_{1})}^{6} {(x - z_{2})}^{6}} = \int \frac{d x}{{(\frac{x - z_{1}}{x - z_{2}})}^{6} {(x - z_{2})}^{12}}$ $c h a n g e m e n t \frac{x - z_{1}}{x - z_{2}} = t g i v e x - z_{1} = t x - {t z}_{2} \Rightarrow$ $(1 - t) x = z_{1} - {t z}_{2} \Rightarrow x = \frac{{t z}_{2} - z_{1}}{t - 1} \Rightarrow$ $d x = \frac{z_{2} (t - 1) - (z_{2} t - z_{1})}{{(t - 1)}^{2}} d t = \frac{z_{1} - z_{2}}{{(t - 1)}^{2}} d t a n d$ $x - z_{2} = \frac{z_{2} t - z_{1}}{t - 1} - z_{2} = \frac{z_{2} t - z_{1} - z_{2} t + z_{2}}{t - 1}$ $= \frac{z_{2} - z_{1}}{(t - 1)} \Rightarrow A = \int \frac{z_{1} - z_{2}}{{(t - 1)}^{2} t^{6} {(\frac{z_{2} - z_{1}}{t - 1})}^{12}} d t$ $= - \frac{1}{{(z_{2} - z_{1})}^{11}} \int \frac{{(t - 1)}^{12}}{{(t - 1)}^{2} t^{6}} d t$ $= \frac{- 1}{{(z_{2} - z_{1})}^{11}} \int \frac{{(t - 1)}^{10}}{t^{6}} d t$ ${(z_{1} - z_{2})}^{11} \times A = \int \frac{\sum_{k = 0}^{10} C_{10}^{k} {(- 1)}^{k} t^{k}}{t^{6}} d t$ $= \int \sum_{k = 0}^{10} {(- 1)}^{k} C_{10}^{k} t^{k - 6} d t$ $= \sum_{k = 0 a n d k \neq 5}^{10} \frac{{(- 1)}^{k}}{k - 5} C_{10}^{k} t^{k - 5} - C_{10}^{5} l n t + C$ $= \sum_{k = 0 a n d k \neq 5}^{10} \frac{{(- 1)}^{k}}{k - 5} C_{10}^{k} {(\frac{x - z_{1}}{x - z_{2}})}^{k - 5} - C_{10}^{5} l n (\frac{x - z_{1}}{x - z_{2}}) + C$ $A = \frac{1}{{(z_{1} - z_{2})}^{11}} \sum_{k = 0 a n d k \neq 5}^{10} \frac{{(- 1)}^{k}}{k - 5} C_{10}^{k} {(\frac{x - z_{1}}{x - z_{2}})}^{k - 5}$ $- \frac{C_{10}^{5}}{{(z_{1} - z_{2})}^{11}} l n (\frac{x - z_{1}}{x - z_{2}}) + C$
	Answered by MJS last updated on 26/Feb/20

	$Ostrogradski$ $Q_{1} (x) = {(x^{2} + x + 1)}^{5}$ $Q_{2} (x) = x^{2} + x + 1$ $P_{1} (x) = \frac{28}{27} (x + \frac{1}{2}) (x^{8} + 4 x^{7} + \frac{21}{6} x^{6} + \frac{35}{2} x^{5} + \frac{223}{10} x^{4} + \frac{201}{10} x^{3} + \frac{1973}{140} x^{2} + \frac{881}{140} x + \frac{297}{140})$ $P_{2} (x) = \frac{28}{27}$ $\int \frac{d x}{{(x^{2} + x + 1)}^{6}} = \frac{P_{1} (x)}{Q_{1} (x)} + \frac{28}{27} \int \frac{d x}{x^{2} + x + 1}$ $now {it}^{'} s simplier than simple$
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