Express-in-partial-fraction-f-x-2x-3-x-2-x-2-1-x-1-x-2-x-1-2-Hence-or-otherwise-show-that-0

Question Number 68675 by Rio Michael last updated on 14/Sep/19

E x p r e s s i n p a r t i a l f r a c t i o n

f (x) \equiv \frac{2 x^{3} + x + 2}{(x^{2} + 1) (x + 1) (x - 2)} x \neq - 1, 2

H e n c e o r o t h e r w i s e s h o w t h a t

\int_{0}^{1} f (x) d x = - \frac{1}{12} [13 l n 2 + π]

Commented by mathmax by abdo last updated on 14/Sep/19

f (x) = \frac{2 x^{3} + x + 2}{(x + 1) (x - 2) (x^{2} + 1)} = \frac{a}{x + 1} + \frac{b}{x - 2} + \frac{c x + d}{x^{2} + 1}

a = {l i m}_{x \to - 1} (x + 1) f (x) = \frac{- 1}{(- 3) (2)} = \frac{1}{6}

b = {l i m}_{x \to 2} (x - 2) f (x) = \frac{16 + 4}{3.5} = \frac{20}{15} = \frac{4}{3}

{l i m}_{x \to + \infty} x f (x) = 2 = a + b + c \Rightarrow c = 2 - a - b = 2 - \frac{1}{6} - \frac{4}{3}

= \frac{12 - 1 - 8}{6} = \frac{1}{2} \Rightarrow f (x) = \frac{1}{6 (x + 1)} + \frac{4}{3 (x - 2)} + \frac{\frac{1}{2} x + d}{x^{2} + 1}

f (0) = - 1 = \frac{1}{6} - \frac{2}{3} + \frac{1}{2} + d = \frac{1 - 4 + 3}{6} + d = d \Rightarrow d = - 1 \Rightarrow

f (x) = \frac{1}{6 (x + 1)} + \frac{4}{3 (x - 2)} + \frac{1}{2} \frac{x - 2}{x^{2} + 1}

\int_{0}^{1} f (x) d x = \frac{1}{6} {[l n ∣ x + 1 ∣]}_{0}^{1} + \frac{4}{3} {[l n ∣ x - 2 ∣]}_{0}^{1} + \frac{1}{4} {[l n (x^{2} + 1)]}_{0}^{1}

- {[a r c t a n x]}_{0}^{1} = \frac{1}{6} l n (2) - \frac{4}{3} l n (2) + \frac{1}{4} l n (2) - \frac{π}{4}

= (\frac{1}{6} - \frac{4}{3} + \frac{1}{4}) l n (2) - \frac{π}{4} = \frac{2 - 16 + 3}{12} l n (2) - \frac{π}{4} \Rightarrow

\int_{0}^{1} f (x) d x = - \frac{11}{12} l n (2) - \frac{π}{4}

Commented by mathmax by abdo last updated on 14/Sep/19

t h e r e i s a e r r o r i n t h e q u e s t i o n!

Commented by Rio Michael last updated on 15/Sep/19

a n y w a y t h a n k s a l o t g u y s i f o u n d m y e r r o r