answer to 25955.we introduce the parametric function F(t) =∫_0 ^∞ ln(1+(1+x^2_ )t)(1+x^2 )^(−1) dx after verifying that F is derivable on[0.∝[ we find ∂F/∂t= ∫_0 ^∞ ( (1+(1+x^2 )t)^(−1) dx ∂F/∂t=1/2 ∫_R (tx^2 +t+1)^(−1) dx we put f(z) =(tz^2 +z+1)^(−1) let find the poles of f..tz^2 +z+1=0 <−> z=+−i((t+1)t^(−1) )^(1/2) and the poles are z_0 =i((t+1)t^(−1) )^(1/2) and z_1 =−i((t+1)t^(−1) )^(1/2) and f(t) =(t(t−z_0 )(t−z_1 ))^(−1) by residus theorem ∫_R f(z)dz =2iπ R(f.z_0 ) =2iπ (t(z_0 −z_1 ))^(−1) =π t^(−1/2) (t+1)^(−1/2 ) −>∂F/∂t =π 2^(−1) t^(−1/2) (1+t)^(−1/2) −>F(t) =π 2^(−1 ) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx +α α=F(0)=0 and F(t) =π2^(−1) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx and by the changement x^(1/2) =u we find F(t) = π ln( t^(1/2) +(1+t)^(1/2) ) so ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx=F(1)=πln(1+2^(1/2) )


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Question Number 25960 by abdo imad last updated on 16/Dec/17

$a n s w e r t o 25955 . w e i n t r o d u c e t h e p a r a m e t r i c f u n c t i o n$ $F (t) = \int_{0}^{\infty} l n (1 + (1 + x^{2_{}}) t) {(1 + x^{2})}^{- 1} d x a f t e r v e r i f y i n g t h a t$ $F i s d e r i v a b l e o n [0 . \propto [w e f i n d \partial F / \partial t = \int_{0}^{\infty} ({(1 + (1 + x^{2}) t)}^{- 1} d x$ $\partial F / \partial t = 1 / 2 \int_{R} {({t x}^{2} + t + 1)}^{- 1} d x w e p u t f (z) = {({t z}^{2} + z + 1)}^{- 1}$ $l e t f i n d t h e p o l e s o f f . . {t z}^{2} + z + 1 = 0 < - > z = + - i {((t + 1) t^{- 1})}^{1 / 2}$ $a n d t h e p o l e s a r e z_{0} = i {((t + 1) t^{- 1})}^{1 / 2} a n d z_{1} = - i {((t + 1) t^{- 1})}^{1 / 2}$ $a n d f (t) = {(t (t - z_{0}) (t - z_{1}))}^{- 1} b y r e s i d u s t h e o r e m$ $\int_{R} f (z) d z = 2 i π R (f . z_{0}) = 2 i π {(t (z_{0} - z_{1}))}^{- 1}$ $= π t^{- 1 / 2} {(t + 1)}^{- 1 / 2} - > \partial F / \partial t = π 2^{- 1} t^{- 1 / 2} {(1 + t)}^{- 1 / 2}$ $- > F (t) = π 2^{- 1} \int_{0}^{t} x^{- 1 / 2} {(1 + x)}^{- 1 / 2} d x + α$ $α = F (0) = 0 a n d F (t) = π 2^{- 1} \int_{0}^{t} x^{- 1 / 2} {(1 + x)}^{- 1 / 2} d x$ $a n d b y t h e c h a n g e m e n t x^{1 / 2} = u w e f i n d$ $F (t) = π l n (t^{1 / 2} + {(1 + t)}^{1 / 2}) s o \int_{0}^{\infty} l n (2 + x^{2}) {(1 + x^{2})}^{- 1} d x = F (1) = π l n (1 + 2^{1 / 2})$
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