let-f-x-0-1-arctan-1-xt-t-2-1-dt-determine-a-explicit-form-for-f-x-2-calculate-0-1-arctan-1-2t-1-t-2-dt-

Question Number 62437 by mathsolverby Abdo last updated on 21/Jun/19

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

d e t e r m i n e a e x p l i c i t f o r m f o r f (x)

2) c a l c u l a t e \int_{0}^{1} \frac{a r c t a n (1 + 2 t)}{1 + t^{2}} d t

Commented by mathmax by abdo last updated on 28/Jun/19

1) w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t}{(1 + x^{2} t^{2}) (t^{2} + 1)} d t =_{x t = u} \int_{0}^{x} \frac{u}{x (1 + u^{2}) (1 + \frac{u^{2}}{x^{2}})} \frac{d u}{x}

= \int_{0}^{x} \frac{u d u}{(1 + u^{2}) (x^{2} + u^{2})} l e t d e c o m p o s e F (u) = \frac{u}{(u^{2} + 1) (u^{2} + x^{2})} \Rightarrow

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

F (- u) = - F (u) \Rightarrow \frac{- a u + b}{u^{2} + 1} + \frac{- c u + d}{u^{2} + x^{2}} = \frac{- a u - b}{u^{2} + 1} + \frac{- c u - d}{u^{2} + x^{2}} \Rightarrow b = d = 0 \Rightarrow

F (u) = \frac{a u}{u^{2} + 1} + \frac{c u}{u^{2} + x^{2}}

{l i m}_{u \to + \infty} u F (u) = 0 = a + c \Rightarrow c = - a \Rightarrow F (u) = \frac{a u}{u^{2} + 1} - \frac{a u}{u^{2} + x^{2}}

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

a (x^{2} - 1) = 1 \Rightarrow a = \frac{1}{x^{2} - 1} (w e s u p p o s e x \neq \bar{+} 1 a n d x \neq 0) \Rightarrow

F (u) = \frac{1}{x^{2} - 1} {\frac{u}{u^{2} + 1} - \frac{u}{u^{2} + x^{2}}} \Rightarrow f^{^{'}} (x) = \int_{0}^{x} F (u) d u

= \frac{1}{x^{2} - 1} {\int_{0}^{x} \frac{u d u}{u^{2} + 1} - \int_{0}^{x} \frac{u d u}{u^{2} + x^{2}}} w e h a v e

\int_{0}^{x} \frac{u d u}{u^{2} + 1} = {[\frac{1}{2} l n (u^{2} + 1)]}_{0}^{x} = \frac{1}{2} l n (x^{2} + 1)

\int_{0}^{x} \frac{u d u}{u^{2} + x^{2}} =_{u = x α} \int_{0}^{1} \frac{x α x d α}{x^{2} (1 + α^{2})} = \int_{0}^{1} \frac{α d α}{1 + α^{2}} = {[\frac{1}{2} l n (1 + α^{2})]}_{0}^{1} = \frac{l n (2)}{2} \Rightarrow

f^{^{'}} (x) = \frac{l n (x^{2} + 1)}{2 (x^{2} - 1)} - \frac{l n (2)}{2 (x^{2} - 1)} \Rightarrow f (x) = \int \frac{l n (1 + x^{2})}{2 (x^{2} - 1)} d x - \frac{l n (2)}{2} \int \frac{d x}{x^{2} - 1} + c

\dots . . b e c o n t i n u e d \dots .