decompose-F-x-x-2-x-4-1-imside-R-x-2-find-the-value-of-2-x-2-x-4-1-

Question Number 33334 by prof Abdo imad last updated on 14/Apr/18

d e c o m p o s e F (x) = \frac{x^{2}}{x^{4} - 1} i m s i d e R (x)

2) f i n d t h e v a l u e o f \int_{2}^{+ \infty} \frac{x^{2}}{x^{4} - 1} .

Commented by prof Abdo imad last updated on 15/Apr/18

2) f i n d t h e v a l u e o f \int_{2}^{+ \infty} \frac{x^{2}}{x^{4} - 1} d x .

Commented by prof Abdo imad last updated on 18/Apr/18

w e h a v e F (x) = \frac{x^{2}}{(x^{2} - 1) (x^{2} + 1)}

= \frac{x^{2} - 1 + 1}{(x^{2} - 1) (x^{2} + 1)} = \frac{1}{x^{2} + 1} + \frac{1}{(x^{2} - 1) (x^{2} + 1)}

l e t d e c o m p o s e g (x) = \frac{1}{(x^{2} - 1) (x^{2} + 1)}

g (x) = \frac{a}{x - 1} + \frac{b}{x + 1} + \frac{c x + d}{x^{2} + 1}

g (- x) = g (x) \Leftrightarrow \frac{- a}{x + 1} + \frac{- b}{x - 1} + \frac{- c x + d}{x^{2} + 1}

= g (x) \Rightarrow b = - a a n d c = 0 \Rightarrow

g (x) = \frac{a}{x - 1} - \frac{a}{x + 1} + \frac{d}{x^{2} + 1}

g (0) = - 1 = - 2 a + d \Rightarrow 2 a - d = 1

g (2) = \frac{1}{3 . 5} = \frac{1}{15} = a - \frac{a}{3} + \frac{d}{5} = \frac{2 a}{3} + \frac{d}{5} \Rightarrow

1 = 10 a + 3 d \Rightarrow 10 a + 3 (2 a - 1) = 1 \Rightarrow

16 a = 4 \Rightarrow a = \frac{1}{4} a n d d = \frac{1}{2} - 1 = - \frac{1}{2}

g (x) = \frac{1}{4 (x - 1)} - \frac{1}{4 (x + 1)} - \frac{1}{2 (x^{2} + 1)} \Rightarrow

F (x) = \frac{1}{x^{2} + 1} + \frac{1}{4 (x - 1)} - \frac{1}{4 (x + 1)} - \frac{1}{2 (x^{2} + 1)}

F (x) = \frac{1}{2 (x^{2} + 1)} + \frac{1}{4 (x - 1)} - \frac{1}{4 (x + 1)}

Commented by math khazana by abdo last updated on 19/Apr/18

\int_{2}^{+ \infty} \frac{x^{2}}{x^{4} - 1} d x = \frac{1}{2} \int_{2}^{+ \infty} \frac{d x}{1 + x^{2}} + \frac{1}{4} \int_{2}^{+ \infty} (\frac{1}{x - 1} - \frac{1}{x + 1}) d x

= \frac{1}{2} {[{a r c t a n x}^{}]}_{2}^{+ \infty} + \frac{1}{4} {[l n ∣ \frac{x - 1}{x + 1} ∣]}_{2}^{+ \infty}

= \frac{1}{2} (\frac{π}{2} - a r c t a n 2) + \frac{1}{4} (- l n (\frac{1}{3}))

= \frac{π}{4} - \frac{1}{2} a r c t a n (2) + \frac{1}{4} l n (3) .