Evaluate : ∫_0 ^1 ((ln (1+x^2 ))/(1+x))d(x).


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Question Number 206579 by York12 last updated on 19/Apr/24

$Evaluate : \underset{0}{\int^{1}} \frac{\ln (1 + x^{2})}{1 + x} d (x) .$
	Answered by Berbere last updated on 19/Apr/24

	$= {[l n (1 + x) l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{l n (1 + x) 2 x}{1 + x^{2}} d x$ $= {l n}^{2} (2) - \int_{0}^{1} \frac{l n (1 + x) 2 x}{1 + x^{2}} = {l n}^{2} (2) - A$ $B = \int_{0}^{1} \frac{l n (1 - x)}{1 + x^{2}} 2 x, A + B = \int_{0}^{1} \frac{2 x l n (1 - x^{2})}{1 + x^{2}} d x$ $= \int_{0}^{1} \frac{l n (1 - x)}{1 + x} d x = \int_{0}^{1} \frac{l n (2) + l n (\frac{1}{2} - \frac{x}{2})}{2 (\frac{1}{2} + \frac{x}{2})} d x$ ${= {l n (2) l n (1 + x)]}_{0}^{1} + \int_{0}^{1} \frac{l n (1 - (\frac{1}{2} + \frac{x}{2})}{\frac{1}{2} + \frac{x}{2}} d (\frac{x}{2} + \frac{1}{2}) = {l n}^{2} (2) - {L i}_{2} (\frac{x + 1}{2})]}_{0}^{1}$ $= {l n}^{2} (2) - {L i}_{2} (1) + {L i}_{2} (\frac{1}{2})$ $B - A = \int_{0}^{1} \frac{2 x l n (\frac{1 - x}{1 + x})}{1 + x^{2}} d x; t = \frac{1 - x}{1 + x}$ $\int_{0}^{1} \frac{2 (1 - t)}{1 + t} . \frac{l n (t)}{\frac{2 (1 + t^{2})}{{(1 + t)}^{2}}} . \frac{2 d t}{{(1 + t)}^{2}} = 2 \int_{0}^{1} \frac{(1 - t) (l n (t)}{(1 + t) (1 + t^{2})}$ $= 2 \int_{0}^{1} \frac{l n (t)}{1 + t} + \frac{- t}{t^{2} + 1} l n (t) = - 2 \int_{0}^{1} \frac{l n (1 + t)}{t} + \int_{0}^{1} \frac{l n (1 + t^{2})}{t} {d t}_{t^{2} \to t}$ $= - 2 \int_{0}^{1} \frac{l n (1 + t)}{t} + \frac{1}{2} \int_{0}^{1} \frac{l n (1 + t)}{t} d t = - \frac{3}{2} \int_{0}^{1} \frac{l n (1 - (- t))}{(- t)} d (- t) =$ ${\frac{3}{2} {L i}_{2} (- t)]}_{0}^{1} = \frac{3 {L i}_{2} (- 1)}{2}$ $A = \frac{1}{2} ((B + A) - (B - A))$ $= \frac{1}{2} ({l n}^{2} (2) - {L i}_{2} (1) + {L i}_{2} (\frac{1}{2}) - \frac{3}{2} {L i}_{2} (- 1))$ $\int_{0}^{1} \frac{l n (1 + x^{2})}{1 + x} d x = {l n}^{2} (2) - A$ $= \frac{{l n}^{2} (2)}{2} + \frac{π^{2}}{48} - \frac{{L i}_{2} (\frac{1}{2})}{2} = \frac{3}{4} {l n}^{2} (2) - \frac{π^{2}}{48}$ ${L i}_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{{l n}^{2} (2)}{2}$
	Commented by York12 last updated on 19/Apr/24

	$Perfect!$ $but there is probably a typo somewhere$ $cause the final answer must be \frac{3}{4} \ln^{2} (2) - \frac{π^{2}}{48}$
	Commented by Berbere last updated on 19/Apr/24

	$y e s y o u r i g h t i c o r r e c t e d i t$
	Commented by York12 last updated on 19/Apr/24

	$thank you$
	Commented by Berbere last updated on 19/Apr/24

	$w i t h e P l e a s u r$
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