find the value of ∫_(−∞) ^(+∞) (x^2 /((1+x +x^2 )^2 ))dx


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Question Number 33736 by prof Abdo imad last updated on 23/Apr/18

$f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(1 + x + x^{2})}^{2}} d x$
Commented by prof Abdo imad last updated on 23/Apr/18

$l e t u s e r e s i d u s t h e o r e m w e h a v e$ $I = \int_{- \infty}^{+ \infty} \frac{x^{2}}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{2}} c h . x + \frac{1}{2} = \frac{\sqrt{3}}{2} t$ $g i v e I = \int_{- \infty}^{+ \infty} \frac{{(\frac{\sqrt{3} t}{2} - \frac{1}{2})}^{2}}{{(\frac{3}{4} t^{2} + \frac{3}{4})}^{2}} \frac{\sqrt{3}}{2} d t$ $= \frac{16}{9} . \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} \frac{3 t^{2} - 2 \sqrt{3} t + 1}{4 {(t^{2} + 1)}^{2}} d t$ $= \frac{2 \sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{3 t^{2} - 2 \sqrt{3} t + 1}{{(t^{2} + 1)}^{2}} d t . l e t c o n s i d e r$ $t h e c o m p l e x f u n c t i o n φ (z) = \frac{3 z^{2} - 2 \sqrt{3} z + 1}{{(z^{2} + 1)}^{2}}$ $t h e p o l e s o f φ a r e i a n d - i (d o u b l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {({(z - i)}^{2} φ (z))}^{^{'}}$ $= {l i m}_{z \to i} {(\frac{3 z^{2} - 2 \sqrt{3} z + 1}{{(z + i)}^{2}})}^{^{'}}$ $= {l i m}_{z \to i} \frac{(6 z - 2 \sqrt{3}) {(z + i)}^{2} - 2 (z + i) (3 z^{2} - 2 \sqrt{3} z + 1)}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(6 z - 2 \sqrt{3}) (z + i) - 2 (3 z^{2} - 2 \sqrt{3} z + 1)}{{(z + i)}^{3}}$ $= \frac{(6 i - 2 \sqrt{3}) (2 i) - 2 (3 i^{2} - 2 \sqrt{3} i + 1)}{{(2 i)}^{3}}$ $= \frac{- 12 - 4 \sqrt{3} i + 6 4 \sqrt{3} i - 2}{- 8 i} = \frac{- 8}{- 8 i} = \frac{1}{i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{1}{i}) = 2 π \Rightarrow$ $I = \frac{2 \sqrt{3}}{9} . (2 π) = \frac{4 π \sqrt{3}}{9}$ $★ I = \frac{4 π \sqrt{3}}{9} ★$
	Answered by MJS last updated on 28/Apr/18

	$O s {trogradski}^{'} s Method :$ $\int \frac{P (x)}{Q (x)} d x = \frac{P_{1} (x)}{Q_{1} (x)} + \int \frac{P_{2} (x)}{Q_{2} (x)} d x$ $Q_{1} (x) = g c d (Q (x); Q^{'} (x))$ $Q_{2} (x) = \frac{Q (x)}{Q_{1} (x)}$ $P (x) = x^{2}$ $Q (x) = {(x^{2} + x + 1)}^{2}$ $Q^{'} (x) = 2 (2 x + 1) (x^{2} + x + 1)$ $g c d (Q (x); Q^{'} (x)) = (x^{2} + x + 1) = Q_{1} (x) = Q_{2} (x)$ $to get P_{1} (x) = k_{1} x + k_{2}, P_{2} (x) = k_{3} x + k_{4}$ $\frac{P (x)}{Q (x)} = {(\frac{P_{1} (x)}{Q_{1} (x)})}^{^{'}} + \frac{P_{2} (x)}{Q_{2} (x)}$ $\frac{x^{2}}{{(x^{2} + x + 1)}^{2}} = \frac{k_{1} (x^{2} + x + 1) - (k_{1} x + k_{2}) (2 x + 1)}{{(x^{2} + x + 1)}^{2}} + \frac{k_{3} x + k_{4}}{x^{2} + x + 1}$ $[multiplicate with {(x^{2} + x + 1)}^{2}]$ $x^{2} = k_{3} x^{3} + (- k_{1} + k_{3} + k_{4}) x^{2} + (- 2 k_{2} + k_{3} + k_{4}) x + (k_{1} - k_{2} + k_{4})$ $k_{3} = 0$ $- k_{1} + k_{3} + k_{4} = 1$ $- 2 k_{2} + k_{3} + k_{4} = 0$ $k_{1} - k_{2} + k_{4} = 0$ $k_{1} = - \frac{1}{3}; k_{2} = \frac{1}{3}; k_{3} = 0; k_{4} = \frac{2}{3}$ $P_{1} (x) = - \frac{x - 1}{3}; P_{2} (x) = \frac{2}{3}$ $\int \frac{P (x)}{Q (x)} d x = \frac{P_{1} (x)}{Q_{1} (x)} + \int \frac{P_{2} (x)}{Q_{2} (x)} d x =$ $= - \frac{x - 1}{3 (x^{2} + x + 1)} + \frac{2}{3} \int \frac{1}{x^{2} + x + 1} d x$ $\frac{1}{x^{2} + x + 1} = \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} = \frac{4}{{(2 x + 1)}^{2} + 3}$ $\frac{8}{3} \int \frac{1}{{(2 x + 1)}^{2} + 3} d x =$ $u = \frac{2 x + 1}{\sqrt{3}} \to d x = \frac{\sqrt{3}}{2} d u$ $x = \frac{u \sqrt{3} - 1}{2}$ $\frac{8}{3} \int \frac{\sqrt{3}}{2 (3 u^{2} + 3)} d u = \frac{4 \sqrt{3}}{9} \int \frac{1}{u^{2} + 1} d u =$ $= \frac{4 \sqrt{3}}{9} \arctan u$ $= \frac{4 \sqrt{3}}{9} \arctan (\frac{\sqrt{3}}{3} (2 x + 1))$ $\int \frac{x^{2}}{{(x^{2} + x + 1)}^{2}} d x = \frac{4 \sqrt{3}}{9} \arctan (\frac{\sqrt{3}}{3} (2 x + 1)) - \frac{x - 1}{3 (x^{2} + x + 1)} + C$ $\underset{- \infty}{\int^{\infty}} \frac{x^{2}}{{(x^{2} + x + 1)}^{2}} d x = \frac{4 \sqrt{3}}{9} π$
	Commented by prof Abdo imad last updated on 23/Apr/18

	$t h a n k y o u s i r f o r t h i s n e w m e t h o d . . .$
	Answered by sma3l2996 last updated on 23/Apr/18

	$I = \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(1 + x + x^{2})}^{2}} d x = \int_{- \infty}^{+ \infty} \frac{x^{2} + x + 1 - (x + 1)}{{(x^{2} + x + 1)}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{1}{x^{2} + x + 1} d x - \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{2 x + 2}{{(x^{2} + x + 1)}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} - \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{2 x + 1}{{(x^{2} + x + 1)}^{2}} d x - \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{2}}$ $= \frac{4}{3} \int_{- \infty}^{+ \infty} \frac{d x}{{(\frac{2 x + 1}{\sqrt{3}})}^{2} + 1} + \frac{1}{2} {[\frac{1}{x^{2} + x + 1}]}_{- \infty}^{+ \infty} - \frac{8}{9} \int_{- \infty}^{+ \infty} \frac{d x}{{({(\frac{2 x + 1}{\sqrt{3}})}^{2} + 1)}^{2}}$ $l e t t = \frac{2 x + 1}{\sqrt{3}} \Rightarrow d x = \frac{\sqrt{3}}{2} d t$ $s o I = \frac{2 \sqrt{3}}{3} \int_{- \infty}^{+ \infty} \frac{d t}{t^{2} + 1} - \frac{4 \sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{2}}$ $l e t t = t a n u \Rightarrow d t = (1 + {t a n}^{2} u) d u$ $I = \frac{2 \sqrt{3}}{3} {[{t a n}^{- 1} (t)]}_{- \infty}^{+ \infty} - \frac{4 \sqrt{3}}{9} \int_{- π / 2}^{π / 2} \frac{d u}{1 + {t a n}^{2} u}$ $= \frac{2 \sqrt{3}}{3} π - \frac{4 \sqrt{3}}{9} \int_{- π / 2}^{π / 2} {c o s}^{2} (u) d u$ $\int_{- π / 2}^{π / 2} {c o s}^{2} (u) d u = \frac{1}{2} \int_{- π / 2}^{π / 2} (c o s (2 u) + 1) d u = \frac{1}{2} {[\frac{1}{2} s i n (2 x) + x]}_{- π / 2}^{π / 2}$ $= \frac{π}{2}$ $I = \frac{2 \sqrt{3}}{3} π - \frac{4 \sqrt{3}}{9} (\frac{π}{2}) = \frac{4 \sqrt{3}}{3} π$
	Commented by MJS last updated on 23/Apr/18

	$just a minor mistake in the last line$ $I = \frac{2 \sqrt{3}}{3} π - \frac{4 \sqrt{3}}{9} (\frac{π}{2}) = \frac{4 \sqrt{3}}{9} π$
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