calculate ∫ (dx/((x^2 +1)(x^2 +2)(x^2 +3))) 1) find the value of ∫


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Question Number 39019 by maxmathsup by imad last updated on 01/Jul/18

$c a l c u l a t e \int \frac{d x}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)}$ $1) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)}$
Commented by math khazana by abdo last updated on 02/Jul/18

$2) \int_{0}^{+ \infty} F (x) d x = [\frac{1}{2} a r c t a n (x) - \frac{1}{2 \sqrt{2}} a r c t a n (\frac{x}{\sqrt{2}})$ ${+ \frac{1}{2 \sqrt{3}} a r c t a n (\frac{x}{\sqrt{3}})]}_{0}^{+ \infty}$ $= \frac{π}{4} - \frac{1}{2 \sqrt{2}} \frac{π}{2} + \frac{1}{2 \sqrt{3}} \frac{π}{2} = \frac{π}{4} - \frac{π}{4 \sqrt{2}} + \frac{π}{4 \sqrt{3}} .$
Commented by math khazana by abdo last updated on 02/Jul/18

$l e t d e c o m p o s e F (x) = \frac{1}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)}$ $= \frac{1}{(u + 1) (u + 2) (u + 3)} = g (u) (u = x^{2})$ $g (u) = \frac{a}{u + 1} + \frac{b}{u + 2} + \frac{c}{u + 3}$ $a = \frac{1}{2}, b = - \frac{1}{2}, c = \frac{1}{(- 2) (- 1)} = \frac{1}{2} \Rightarrow$ $F (x) = \frac{1}{2 (x^{2} + 1)} - \frac{1}{2 (x^{2} + 2)} + \frac{1}{2 (x^{2} + 3)}$ $\int F (x) d x = \frac{1}{2} \int \frac{d x}{x^{2} + 1} - \frac{1}{2} \int \frac{d x}{x^{2} + 2} + \frac{1}{2} \int \frac{d x}{x^{2} + 3} + c$ $\int \frac{d x}{x^{2} + 1} = a r c t a n x + c_{1}$ $\int \frac{d x}{x^{2} + 2} =_{x = \sqrt{2} u} \int \frac{\sqrt{2} d u}{2 (1 + u^{2})} = \frac{1}{\sqrt{2}} a r c t a n (\frac{x}{\sqrt{2}}) + c_{2}$ $\int \frac{d x}{x^{2} + 3} = \frac{1}{\sqrt{3}} a r c t a n (\frac{x}{\sqrt{3}}) + c_{3} \Rightarrow$ $\int F (x) d x = \frac{1}{2} a r c t a n (x) - \frac{1}{2 \sqrt{2}} a r c t a n (\frac{x}{\sqrt{2}})$ $+ \frac{1}{2 \sqrt{3}} a r c t a n (\frac{x}{\sqrt{3}}) + c$
Commented by math khazana by abdo last updated on 03/Jul/18
$2) Residus method let I=∫_0 ^∞ (dx/((x^2 +1)(x^(2 ) +2)(x^2 +3))) 2I = ∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +2)(x^2 +3))) let ϕ(z) = (1/((z^2 +1)(z^2 +2)(z^2 +3))) the poles of ϕ are i,−i,i(√2),−i(√2),i(3)^(1/�) ,−i(√3) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,i(√2))+Res(ϕ^� ,−i(√3))} ϕ(z) =(1/((z−i)(z+i)(z−i(√2))(z+i(√2))(z+i(√3))(z−i(√3)))) Res(ϕ,i) = (1/((2i)(1)2)) =(1/(4i)) Res(ϕ,i(√2)) = (1/((2i(√2))(−1)(1))) =−(1/(2i(√2))) Res(ϕ,i(√3)) = (1/((2i(√3))(−2)(−1))) =(1/(4i(√3))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(4i)) −(1/(2i(√2))) +(1/(4i(√3)))} = (π/2) −(π/(√2)) + (π/(2(√3))) ⇒ I = (π/4) −(π/(2(√2))) +(π/(4(√3))) .$
$2) R e s i d u s m e t h o d l e t I = \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)}$ $2 I = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)}$ $l e t φ (z) = \frac{1}{(z^{2} + 1) (z^{2} + 2) (z^{2} + 3)}$ $t h e p o l e s o f φ a r e i, - i, i \sqrt{2}, - i \sqrt{2}, i \sqrt[]{3}, - i \sqrt{3}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, i \sqrt{2}) + R e s (\bar{φ}, - i \sqrt{3})}$ $φ (z) = \frac{1}{(z - i) (z + i) (z - i \sqrt{2}) (z + i \sqrt{2}) (z + i \sqrt{3}) (z - i \sqrt{3})}$ $R e s (φ, i) = \frac{1}{(2 i) (1) 2} = \frac{1}{4 i}$ $R e s (φ, i \sqrt{2}) = \frac{1}{(2 i \sqrt{2}) (- 1) (1)} = - \frac{1}{2 i \sqrt{2}}$ $R e s (φ, i \sqrt{3}) = \frac{1}{(2 i \sqrt{3}) (- 2) (- 1)} = \frac{1}{4 i \sqrt{3}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{4 i} - \frac{1}{2 i \sqrt{2}} + \frac{1}{4 i \sqrt{3}}}$ $= \frac{π}{2} - \frac{π}{\sqrt{2}} + \frac{π}{2 \sqrt{3}} \Rightarrow$ $I = \frac{π}{4} - \frac{π}{2 \sqrt{2}} + \frac{π}{4 \sqrt{3}} .$
	Answered by behi83417@gmail.com last updated on 03/Jul/18

	$I = \frac{1}{3} \int \frac{2 (x^{2} + 3) - (x^{2} + 1) - (x^{2} + 2)}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} d x =$ $= \frac{1}{3} \int [\frac{2}{(x^{2} + 1) (x^{2} + 2)} - \frac{1}{(x^{2} + 2) (x^{2} + 3)} - \frac{1}{(x^{2} + 1) (x^{2} + 3)}] d x =$ $= \frac{1}{3} \int [2 (\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 2}) + \frac{1}{x^{2} + 3} - \frac{1}{x^{2} + 2} - \frac{1}{2} (\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 3})] d x =$ $= \frac{1}{3} \int [\frac{3}{2 (x^{2} + 1)} - \frac{3}{2 (x^{2} + 1)} + \frac{3}{2 (x^{2} + 3)}] d x =$ $= \frac{1}{2} {t g}^{- 1} x - \frac{1}{2 \sqrt{2}} {t g}^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{2 \sqrt{3}} {t g}^{- 1} \frac{x}{\sqrt{3}} + c o n s t .$ $2) \underset{0}{\int^{\infty}} f (x) d x = (\frac{1}{2} - \frac{1}{2 \sqrt{2}} + \frac{1}{2 \sqrt{3}}) . \frac{π}{2} =$ $= \frac{π}{4} - \frac{π}{4 \sqrt{2}} + \frac{π}{4 \sqrt{3}} . ◼$
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