calculate-f-a-1-a-x-2-arctan-x-a-x-dx-with-a-real-

Question Number 60691 by maxmathsup by imad last updated on 24/May/19

c a l c u l a t e f (a) = \int (1 - \frac{a}{x^{2}}) a r c t a n (x + \frac{a}{x}) d x w i t h a r e a l .

Commented by maxmathsup by imad last updated on 27/May/19

b y p a r t s u^{^{'}} = 1 - \frac{a}{x^{2}} a n d v = a r c t a n (x + \frac{a}{x}) \Rightarrow

f (a) = (x + \frac{a}{x}) a r c t a n (x + \frac{a}{x}) - \int (x + \frac{a}{x}) \frac{1 - \frac{a}{x^{2}}}{1 + {(x + \frac{a}{x})}^{2}} d x

= (x + \frac{a}{x}) a r c t a n (x + \frac{a}{x}) - \int (x + \frac{a}{x}) \frac{x^{2} - a}{x^{2} + {(x^{2} + a^{2})}^{2}} d x b u t

\int (x + \frac{a}{x}) \frac{x^{2} - a}{x^{2} + {(x^{2} + a^{2})}^{2}} = \int \frac{x^{4} - a^{2}}{x (x^{2} + x^{4} + 2 x^{2} a^{2} + a^{4})} d x

= \int \frac{x^{4} - a^{2}}{x (x^{4} + (2 a^{2} + 1) x^{2} + a^{4})} d x l e t F (x) = \frac{x^{4} - a^{2}}{x (x^{4} + (2 a^{2} + 1) x^{2} + a^{4})}

p o l e s o f F ?

r o o t s o f x^{4} + (2 a^{2} + 1) x^{2} + a^{4} = 0 \Rightarrow t^{2} + (2 a^{2} + 1) t + a^{4} = 0 (t = x^{2})

Δ = {(2 a^{2} + 1)}^{2} - 4 a^{4} = 4 a^{4} + 4 a^{2} + 1 - 4 a^{4} = 4 a^{2} + 1 \Rightarrow

t_{1} = \frac{- 2 a^{2} - 1 + \sqrt{4 a^{2} + 1}}{2} a n d t_{2} = \frac{- 2 a^{2} - 1 - \sqrt{4 a^{2} + 1}}{2}

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

t h e d e c o m p o s i t i o n o f F (x) i s a t f o r m

F (x) = \frac{a}{x} + \frac{b x + c}{x^{2} + \frac{2 a^{2} + 1 - \sqrt{4 a^{2} + 1}}{2}} + \frac{d x + e}{x^{2} + \frac{2 a^{2} + 1 + \sqrt{4 a^{2} + 1}}{2}}

a = {l i m}_{x \to a} x F (x) = \frac{a^{4} - a^{2}}{(a^{2} - t_{1}) (a^{2} - t_{2})}

{l i m}_{x \to + \infty} x F (x) = 1 = a + b + d \Rightarrow b + d = - a \dots . b e c o n t i n u e d \dots .