1-calculate-f-a-0-a-2x-1-x-2-x-3-x-2-1-dx-1-calculate-f-1-and-f-2-

Question Number 57225 by maxmathsup by imad last updated on 31/Mar/19

1) c a l c u l a t e f (a) = \int_{0}^{a} \frac{2 x - 1}{(x^{2} - x + 3) (x^{2} + 1)} d x

1) c a l c u l a t e f (1) a n d f (2)

Commented by maxmathsup by imad last updated on 04/Apr/19

l e t d e c o m p o s e F (x) = \frac{2 x - 1}{(x^{2} - x + 3) (x^{2} + 1)} \Rightarrow

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

{a x}^{3} + a x + {b x}^{2} + b + {c x}^{3} - {c x}^{2} + 3 c x + {d x}^{2} - d x + 3 d = 2 x - 1 \Rightarrow

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

a + c = 0 a n d b - c + d = 0 a n d a + 3 c - d = 2 a n d 3 d = - 1 \Rightarrow d = - \frac{1}{3}

c = - a \Rightarrow a - 3 a - d = 2 \Rightarrow - 2 a = d + 2 = - \frac{1}{3} + 2 = \frac{5}{3} \Rightarrow a = - \frac{5}{6}

c = \frac{5}{6}, b = c - d = \frac{5}{6} + \frac{1}{3} = \frac{7}{6} \Rightarrow

F (x) = \frac{- \frac{5}{6} x + \frac{7}{6}}{x^{2} - x + 3} + \frac{\frac{5}{6} x - \frac{2}{6}}{x^{2} + 1} = - \frac{1}{6} \frac{x - 7}{x^{2} - x + 3} + \frac{1}{6} \frac{5 x - 2}{x^{2} + 1} \Rightarrow

\int F (x) d x = - \frac{1}{12} \int \frac{2 x - 14}{x^{2} - x + 3} d x + \frac{5}{12} \int \frac{2 x}{x^{2} + 1} - \frac{1}{3} \int \frac{d x}{x^{2} + 1}

= - \frac{1}{12} l n (x^{2} - x + 3) + \frac{13}{12} \int \frac{d x}{x^{2} - x + 3} + \frac{5}{12} l n (x^{2} + 1) - \frac{1}{3} a r c t a n (x) + c

b u t \int \frac{d x}{x^{2} - x + 3} = \int \frac{d x}{x^{2} - 2 \frac{x}{2} + \frac{1}{4} + 3 - \frac{1}{4}} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{11}{4}}

=_{x - \frac{1}{2} = \frac{\sqrt{11}}{2} u} \frac{4}{11} \int \frac{1}{1 + u^{2}} \frac{\sqrt{11}}{2} d u = \frac{2}{\sqrt{11}} a r c t a n (\frac{2 x - 1}{\sqrt{11}}) \Rightarrow

\int F (x) d x = - \frac{1}{12} l n (x^{2} - x + 3) + \frac{13}{6 \sqrt{11}} a r c t a n (\frac{2 x - 1}{\sqrt{11}}) + \frac{5}{12} l n (x^{2} + 1) - \frac{1}{3} a r c t a n (x) + c

\Rightarrow f (a) = \int_{0}^{a} F (x) d x = {[- \frac{1}{12} l n (x^{2} - x + 3) + \frac{13}{6 \sqrt{11}} a r c t a n (\frac{2 x - 1}{\sqrt{11}}) + \frac{5}{12} l n (x^{2} + 1) - \frac{1}{3} a r c t a n (x)]}_{0}^{a}

= - \frac{1}{12} l n (a^{2} - a + 3) + \frac{13}{6 \sqrt{11}} a r c t a n (\frac{2 a - 1}{\sqrt{11}}) + \frac{5}{12} l n (a^{2} + 1) - \frac{1}{3} a r c t a n (a)

+ \frac{1}{12} l n (3) + \frac{13}{6 \sqrt{11}} a r c t a n (\frac{1}{\sqrt{11}})

Commented by maxmathsup by imad last updated on 05/Apr/19

2) f (1) = - \frac{1}{12} l n (3) + \frac{13}{6 \sqrt{11}} a r c t a n (\frac{1}{\sqrt{11}}) + \frac{5}{12} l n (2) - \frac{π}{12} + \frac{1}{12} l n (3) + \frac{13}{6 \sqrt{11}} a r c t a n (\frac{1}{\sqrt{11}})

f (1) = \frac{13}{3 \sqrt{11}} a r c t a n (\frac{1}{\sqrt{11}}) + \frac{5}{12} l n (2) - \frac{π}{12} .