calculate ∫_0 ^∞ (dx/((x+1)(x+2)(x+3)))


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Question Number 44302 by abdo.msup.com last updated on 26/Sep/18

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{(x + 1) (x + 2) (x + 3)}$
Commented by maxmathsup by imad last updated on 28/Sep/18
$let I =∫_0 ^∞ (dx/((x+1)(x+2)(x+3))) let decompose F(x)=(1/((x+1)(x+2)(x+3))) F(x)=(a/(x+1)) +(b/(x+2)) +(c/(x+3)) a =lim_(x→−1) (x+1)F(x)=(1/2) b=lim_(x→−2) (x+2)F(x) = −1 c =lim_(x→−3) (x+3)F(x)=(1/2) ⇒F(x)= (1/(2(x+1))) −(1/((x+2))) +(1/(2(x+3))) ⇒∫_0 ^∞ F(x)dx =∫_0 ^∞ { (1/(2(x+1))) −(1/(x+2)) +(1/(2(x+3)))}dx =[(1/2)ln∣x+1∣−ln∣x+2∣+(1/2)ln∣x+3∣]_0 ^(+∞) =[ln(((√((x+1)(x+3)))/(x+2))]_0 ^(+∞) −ln(((√3)/2)) =ln(2)−(1/2)ln(3) ⇒ I =ln(2)−((ln(3))/2) .$
$l e t I = \int_{0}^{\infty} \frac{d x}{(x + 1) (x + 2) (x + 3)} l e t d e c o m p o s e F (x) = \frac{1}{(x + 1) (x + 2) (x + 3)}$ $F (x) = \frac{a}{x + 1} + \frac{b}{x + 2} + \frac{c}{x + 3}$ $a = {l i m}_{x \to - 1} (x + 1) F (x) = \frac{1}{2}$ $b = {l i m}_{x \to - 2} (x + 2) F (x) = - 1$ $c = {l i m}_{x \to - 3} (x + 3) F (x) = \frac{1}{2} \Rightarrow F (x) = \frac{1}{2 (x + 1)} - \frac{1}{(x + 2)} + \frac{1}{2 (x + 3)}$ $\Rightarrow \int_{0}^{\infty} F (x) d x = \int_{0}^{\infty} {\frac{1}{2 (x + 1)} - \frac{1}{x + 2} + \frac{1}{2 (x + 3)}} d x$ $= {[\frac{1}{2} l n ∣ x + 1 ∣ - l n ∣ x + 2 ∣ + \frac{1}{2} l n ∣ x + 3 ∣]}_{0}^{+ \infty} = [l n {(\frac{\sqrt{(x + 1) (x + 3)}}{x + 2}]}_{0}^{+ \infty}$ $- l n (\frac{\sqrt{3}}{2}) = l n (2) - \frac{1}{2} l n (3) \Rightarrow I = l n (2) - \frac{l n (3)}{2} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Sep/18
	$(1/((x+1)(x+2)(x+3)))=(a/(x+1))+(b/(x+2))+(c/(x+3)) 1=a(x+2)(x+3)+b(x+1)(x+3)+c(x+1)(x+2) put x+1=0 x=−1 1=a(−1+2)(−1+3) a=(1/2) x+2=0 1=b(−2+1)(−2+3) b=−1 x+3=0 1=c(−3+1)(−3+2) c=(1/2) ∫_0 ^∞ ((1/2)/(x+1))dx+∫_0 ^∞ ((−1)/(x+2))dx+∫_0 ^∞ ((1/2)/(x+3))dx {(1/2)ln∣x+1∣−ln∣x+2∣+(1/2)ln∣x+3∣}_0 ^∞ ={ln∣((√((x+1)(x+3)))/(x+2))∣}_0 ^∞ ={ln∣((x.(√((1+(1/x))(1+(3/x)))))/(x(1+(2/x))))∣}_0 ^∞ ={ln1−ln∣((√(1×3))/2)∣} =0−{(1/2)ln3−ln2} =ln2−(1/2)ln3$
	$\frac{1}{(x + 1) (x + 2) (x + 3)} = \frac{a}{x + 1} + \frac{b}{x + 2} + \frac{c}{x + 3}$ $1 = a (x + 2) (x + 3) + b (x + 1) (x + 3) + c (x + 1) (x + 2)$ $p u t x + 1 = 0 x = - 1$ $1 = a (- 1 + 2) (- 1 + 3) a = \frac{1}{2}$ $x + 2 = 0$ $1 = b (- 2 + 1) (- 2 + 3) b = - 1$ $x + 3 = 0$ $1 = c (- 3 + 1) (- 3 + 2) c = \frac{1}{2}$ $\int_{0}^{\infty} \frac{\frac{1}{2}}{x + 1} d x + \int_{0}^{\infty} \frac{- 1}{x + 2} d x + \int_{0}^{\infty} \frac{\frac{1}{2}}{x + 3} d x$ ${\frac{1}{2} l n ∣ x + 1 ∣ - l n ∣ x + 2 ∣ + \frac{1}{2} l n ∣ x + 3 ∣}_{0}^{\infty}$ $= {l n ∣ \frac{\sqrt{(x + 1) (x + 3)}}{x + 2} ∣}_{0}^{\infty}$ $= {l n ∣ \frac{x . \sqrt{(1 + \frac{1}{x}) (1 + \frac{3}{x})}}{x (1 + \frac{2}{x})} ∣}_{0}^{\infty}$ $= {l n 1 - l n ∣ \frac{\sqrt{1 \times 3}}{2} ∣}$ $= 0 - {\frac{1}{2} l n 3 - l n 2}$ $= l n 2 - \frac{1}{2} l n 3$
	Commented by maxmathsup by imad last updated on 28/Sep/18

	$y o u r a n s w e r i s c o r r e c t s i r T a n m a y t h a n k s . . .$
	Answered by ajfour last updated on 28/Sep/18

	$l e t x + 2 = z$ $\Rightarrow I = \int_{2}^{\infty} \frac{d z}{z (z^{2} - 1)}$ $= \frac{1}{2} \int \frac{d z}{z + 1} - \int \frac{d z}{z} + \frac{1}{2} \int \frac{d z}{z - 1}$ $= \frac{1}{2} \ln ∣ \frac{z^{2} - 1}{z^{2}} ∣_{2}^{\infty}$ $= \frac{1}{2} \ln \frac{4}{3} = \ln 2 - \frac{1}{2} \ln 3 .$
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