find ∫_0 ^1 arctan((2/(1+x)))dx


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Question Number 61039 by mathsolverby Abdo last updated on 28/May/19

$f i n d \int_{0}^{1} a r c t a n (\frac{2}{1 + x}) d x$
Commented by maxmathsup by imad last updated on 29/May/19
$let A =∫_0 ^1 arctn((2/(1+x)))dx by parts u^′ =1 and v=arctan((2/(1+x))) ⇒ v^′ =(((−2)/((1+x)^2 ))/(1+((2/(1+x)))^2 )) =((−2)/((1+x^2 ){1+(4/((1+x)^2 ))})) =((−2)/(1+x^2 +4)) =((−2)/(x^2 +5)) ⇒ A =[x arctan((2/(1+x)))]_0 ^1 −∫_0 ^1 ((−2x)/(x^2 +5)) dx =(π/4) +∫_0 ^1 ((2x)/(x^2 +5)) dx =(π/4) +[ln(x^2 +5)]_0 ^1 =(π/4) +ln(6)−ln(5) .$
$l e t A = \int_{0}^{1} a r c t n (\frac{2}{1 + x}) d x b y p a r t s u^{^{'}} = 1 a n d v = a r c t a n (\frac{2}{1 + x}) \Rightarrow$ $v^{^{'}} = \frac{\frac{- 2}{{(1 + x)}^{2}}}{1 + {(\frac{2}{1 + x})}^{2}} = \frac{- 2}{(1 + x^{2}) {1 + \frac{4}{{(1 + x)}^{2}}}} = \frac{- 2}{1 + x^{2} + 4} = \frac{- 2}{x^{2} + 5} \Rightarrow$ $A = {[x a r c t a n (\frac{2}{1 + x})]}_{0}^{1} - \int_{0}^{1} \frac{- 2 x}{x^{2} + 5} d x = \frac{π}{4} + \int_{0}^{1} \frac{2 x}{x^{2} + 5} d x$ $= \frac{π}{4} + {[l n (x^{2} + 5)]}_{0}^{1} = \frac{π}{4} + l n (6) - l n (5) .$
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