find ∫_(−∞) ^(+∞) (((x^2 +2)dx)/(x^4 +8x^2 −16x +20)) .


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Question Number 29854 by abdo imad last updated on 13/Feb/18

$f i n d \int_{- \infty}^{+ \infty} \frac{(x^{2} + 2) d x}{x^{4} + 8 x^{2} - 16 x + 20} .$
Commented by abdo imad last updated on 18/Feb/18

$w e h a v e p r o v e d t h a t \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 i x + 2 - 4 i} = - \frac{2 π}{3} - i \frac{π}{3} b u t$ $\frac{1}{x^{2} + 2 i x + 2 - 4 i} = \frac{1}{x^{2} + 2 + i (2 x - 4)}$ $= \frac{x^{2} + 2 - i (2 x - 4)}{{(x^{2} + 2)}^{2} + {(2 x - 4)}^{2}} = \frac{x^{2} + 2}{x^{4} + 4 x^{2} + 4 + 4 x^{2} - 16 x + 16}$ $- i \frac{2 x - 4}{x^{4} + 4 x^{2} + 4 + 4 x^{2} - 16 x + 16}$ $= \frac{x^{2} + 2}{x^{4} + 8 x^{2} - 16 x + 20} - i \frac{2 x - 4}{x^{4} + 8 x^{2} - 16 x + 20} \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{x^{2} + 2}{x^{4} + 8 x^{2} - 16 x + 20} = \frac{- 2 π}{3} a n d$ $\int_{- \infty}^{+ \infty} \frac{2 x - 4}{x^{4} + 8 x^{2} - 16 x + 20} = \frac{π}{3} .$
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