let f(x) =∫_0 ^1 ((arctan(1+xt))/(t^2 +1))dt determine a explicit form for f(x) 2)calculate ∫


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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
$1) we have f^′ (x) =∫_0 ^1 (t/((1+x^2 t^2 )(t^2 +1)))dt =_(xt =u) ∫_0 ^x (u/(x(1+u^2 )(1+(u^2 /x^2 )))) (du/x) =∫_0 ^x ((udu)/((1+u^2 )(x^2 +u^2 ))) let decompose F(u) =(u/((u^2 +1)(u^2 +x^2 ))) ⇒ F(u) =((au +b)/(u^2 +1)) +((cu +d)/(u^2 +x^2 )) F(−u) =−F(u)⇒((−au +b)/(u^2 +1)) +((−cu +d)/(u^2 +x^2 )) =((−au−b)/(u^2 +1)) +((−cu−d)/(u^2 +x^2 )) ⇒b=d =0 ⇒ F(u) =((au)/(u^2 +1)) +((cu)/(u^2 +x^2 )) lim_(u→+∞) uF(u) =0 =a+c ⇒c=−a ⇒F(u) =((au)/(u^2 +1)) −((au)/(u^2 +x^2 )) F(1) =(1/(2(x^2 +1))) =(a/2) −(a/(x^2 +1)) ⇒((ax^2 +a−2a)/(2(x^(2 ) +1))) =(1/(2(x^2 +1))) ⇒ax^2 −a =1 ⇒ a(x^2 −1) =1 ⇒ a =(1/(x^2 −1)) (we suppose x≠+^− 1 and x≠0) ⇒ F(u) =(1/(x^2 −1)){(u/(u^2 +1)) −(u/(u^2 +x^2 ))}⇒f^′ (x) =∫_0 ^x F(u)du =(1/(x^2 −1)) { ∫_0 ^x ((udu)/(u^2 +1)) −∫_0 ^x ((udu)/(u^2 +x^2 ))} we have ∫_0 ^x ((udu)/(u^2 +1)) =[(1/2)ln(u^2 +1)]_0 ^x =(1/2)ln(x^2 +1) ∫_0 ^x ((udu)/(u^2 +x^2 )) =_(u =xα) ∫_0 ^1 ((xα xdα)/(x^2 (1+α^2 ))) =∫_0 ^1 ((αdα)/(1+α^2 )) =[(1/2)ln(1+α^2 )]_0 ^1 =((ln(2))/2) ⇒ f^′ (x) =((ln(x^2 +1))/(2(x^2 −1))) −((ln(2))/(2(x^2 −1))) ⇒f(x) =∫ ((ln(1+x^2 ))/(2(x^2 −1)))dx −((ln(2))/2) ∫ (dx/(x^2 −1)) +c .....be continued....$
$1) w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t}{(1 + x^{2} t^{2}) (t^{2} + 1)} d t =_{x t = u} \int_{0}^{x} \frac{u}{x (1 + u^{2}) (1 + \frac{u^{2}}{x^{2}})} \frac{d u}{x}$ $= \int_{0}^{x} \frac{u d u}{(1 + u^{2}) (x^{2} + u^{2})} l e t d e c o m p o s e F (u) = \frac{u}{(u^{2} + 1) (u^{2} + x^{2})} \Rightarrow$ $F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + x^{2}}$ $F (- u) = - F (u) \Rightarrow \frac{- a u + b}{u^{2} + 1} + \frac{- c u + d}{u^{2} + x^{2}} = \frac{- a u - b}{u^{2} + 1} + \frac{- c u - d}{u^{2} + x^{2}} \Rightarrow b = d = 0 \Rightarrow$ $F (u) = \frac{a u}{u^{2} + 1} + \frac{c u}{u^{2} + x^{2}}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + c \Rightarrow c = - a \Rightarrow F (u) = \frac{a u}{u^{2} + 1} - \frac{a u}{u^{2} + x^{2}}$ $F (1) = \frac{1}{2 (x^{2} + 1)} = \frac{a}{2} - \frac{a}{x^{2} + 1} \Rightarrow \frac{{a x}^{2} + a - 2 a}{2 (x^{2} + 1)} = \frac{1}{2 (x^{2} + 1)} \Rightarrow {a x}^{2} - a = 1 \Rightarrow$ $a (x^{2} - 1) = 1 \Rightarrow a = \frac{1}{x^{2} - 1} (w e s u p p o s e x \neq \bar{+} 1 a n d x \neq 0) \Rightarrow$ $F (u) = \frac{1}{x^{2} - 1} {\frac{u}{u^{2} + 1} - \frac{u}{u^{2} + x^{2}}} \Rightarrow f^{^{'}} (x) = \int_{0}^{x} F (u) d u$ $= \frac{1}{x^{2} - 1} {\int_{0}^{x} \frac{u d u}{u^{2} + 1} - \int_{0}^{x} \frac{u d u}{u^{2} + x^{2}}} w e h a v e$ $\int_{0}^{x} \frac{u d u}{u^{2} + 1} = {[\frac{1}{2} l n (u^{2} + 1)]}_{0}^{x} = \frac{1}{2} l n (x^{2} + 1)$ $\int_{0}^{x} \frac{u d u}{u^{2} + x^{2}} =_{u = x α} \int_{0}^{1} \frac{x α x d α}{x^{2} (1 + α^{2})} = \int_{0}^{1} \frac{α d α}{1 + α^{2}} = {[\frac{1}{2} l n (1 + α^{2})]}_{0}^{1} = \frac{l n (2)}{2} \Rightarrow$ $f^{^{'}} (x) = \frac{l n (x^{2} + 1)}{2 (x^{2} - 1)} - \frac{l n (2)}{2 (x^{2} - 1)} \Rightarrow f (x) = \int \frac{l n (1 + x^{2})}{2 (x^{2} - 1)} d x - \frac{l n (2)}{2} \int \frac{d x}{x^{2} - 1} + c$ $. . . . . b e c o n t i n u e d . . . .$
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