calculate-0-dx-x-1-x-2-x-3-

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

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

\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 \dots

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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