calculate ∫_0 ^1 arctan(1+x+x^2 )dx


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Question Number 59274 by Mr X pcx last updated on 07/May/19

$c a l c u l a t e \int_{0}^{1} a r c t a n (1 + x + x^{2}) d x$
Commented by maxmathsup by imad last updated on 18/May/19

$w e h a v e I = \int_{0}^{1} a r c t a n (1 + x + x^{2}) = \int_{0}^{1} a r c o t a n (\frac{1}{1 + x + x^{2}})$ $= \int_{0}^{1} a r c o t a n (\frac{x + 1 - x}{1 + (x + 1) x}) d x l e t x + 1 = t a n α a n d x = t a n β \Rightarrow$ $I = \int_{0}^{1} a r c o t a n (\frac{t a n α - t a n β}{1 + t a n α . t a n β}) d x = \int_{0}^{1} a r c o t a n (t a n (α - β)$ $= \int_{0}^{1} a r c o t a n (c o t a n (\frac{π}{2} - α + β)) d x = \int_{0}^{1} (\frac{π}{2} - α + β) d x$ $= \frac{π}{2} - \int_{0}^{1} a r c t a n (x + 1) d x + \int_{0}^{1} a r c t a n x d x b y p a r t s$ $\int_{0}^{1} a r c t a n x d x = {[x a r c t a n x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^{2}} d x = \frac{π}{4} - {[\frac{1}{2} l n (1 + x^{2})]}_{0}^{1}$ $= \frac{π}{4} - \frac{l n (2)}{2}$ $\int_{0}^{1} a r c t a n (x + 1) d x = \int_{1}^{2} a r c t a n (x) d x = {[x a r c t a n x]}_{1}^{2} - \int_{1}^{2} \frac{x}{1 + x^{2}} d x$ $= 2 a r c t a n 2 - \frac{π}{4} - \frac{1}{2} {[l n (1 + x^{2})]}_{1}^{2} = 2 a r c t a n (2) - \frac{π}{4} - \frac{1}{2} (l n (5) - l n (2)) \Rightarrow$ $I = \frac{π}{2} - 2 a r c t a n (2) + \frac{π}{4} + \frac{1}{2} l n (5) - \frac{1}{2} l n (2) + \frac{. π}{4} - \frac{l n (2)}{2}$ $I = π - 2 a r c t a n (2) + \frac{l n (5) - l n (2)}{2} .$
Commented by maxmathsup by imad last updated on 18/May/19

$e r r o r a t f i n a l l i n e I = π - 2 a r c t a n (2) + \frac{l n (5)}{2} - l n (2) .$
	Answered by tanmay last updated on 07/May/19
	$∫_0 ^1 tan^(−1) (1+x+x^2 )dx ∫_0 ^1 cot^(−1) ((1/(1+x(x+1))))dx x+1=tana x=tanb (1/(1+x(x+1)))=((tana−tanb)/(1+tanatanb))=tan(a−b)=cot[(π/2)−(a−b)] cot^(−1) [cot{(π/2)−(a−b)}] =(π/2)−a+b =(π/2)−tan^− (x+1)+tan^(−1) x ∫_0 ^1 (π/2)dx−∫_0 ^1 tan^(−1) (x+1)dx+∫_0 ^1 tan^(−1) xdx now ∫_0 ^1 (π/2)dx=(π/2) ∫_0 ^1 tan^(−1) (x+1)dx ★∫tan^(−1) (x+1)dx =tan^(−1) (x+1)×x−∫(x/(1+(1+x)^2 ))dx =xtan^(−1) (x+1)−(1/2)∫((2x+2−2)/(x^2 +2x+2)) =xtan^(−1) (x+1)−(1/2)∫((d(x^2 +2x+2))/(x^2 +2x+2))+∫(dx/(1+(x+1)^2 )) =xtan^(−1) (x+1)−(1/2)ln(x^2 +2x+2)+tan^(−1) (x+1) so ∣xtan^(−1) (x+1)−(1/2)ln(x^2 +2x+2)+tan^(−1) (x+1)∣_0 ^1 =■tan^(−1) (2)−(1/2)ln(5)+tan^− (2)+(1/2)ln2−(π/4)■ ∫_0 ^1 tan^(−1) xdx ∫tan^(−1) xdx =xtan^(−1) x−∫(x/(1+x^2 ))dx =xtan^(−1) x−(1/2)ln(1+x^2 ) so∣xtan^(−1) x−(1/2)ln(1+x^2 )∣_0 ^1 =tan^(−1) (1)−(1/2)ln2 =(π/4)−(1/2)ln2 so answer is ∫_0 ^1 (π/2)dx+∫_0 ^1 tan^(−1) xdx−∫_0 ^1 tan^(−1) (x+1)dx =(π/2)+(π/4)−(1/2)ln2−2tan^(−1) (2)+(1/2)ln5−(1/2)ln2+(π/4) =π−ln2+(1/2)ln5−2tan^(−1) (2) pls go through...$
	$\int_{0}^{1} {t a n}^{- 1} (1 + x + x^{2}) d x$ $\int_{0}^{1} {c o t}^{- 1} (\frac{1}{1 + x (x + 1)}) d x$ $x + 1 = t a n a x = t a n b$ $\frac{1}{1 + x (x + 1)} = \frac{t a n a - t a n b}{1 + t a n a t a n b} = t a n (a - b) = c o t [\frac{π}{2} - (a - b)]$ ${c o t}^{- 1} [c o t {\frac{π}{2} - (a - b)}]$ $= \frac{π}{2} - a + b$ $= \frac{π}{2} - {t a n}^{-} (x + 1) + {t a n}^{- 1} x$ $\int_{0}^{1} \frac{π}{2} d x - \int_{0}^{1} {t a n}^{- 1} (x + 1) d x + \int_{0}^{1} {t a n}^{- 1} x d x$ $n o w \int_{0}^{1} \frac{π}{2} d x = \frac{π}{2}$ $\int_{0}^{1} {t a n}^{- 1} (x + 1) d x$ $★ \int {t a n}^{- 1} (x + 1) d x$ $= {t a n}^{- 1} (x + 1) \times x - \int \frac{x}{1 + {(1 + x)}^{2}} d x$ $= {x t a n}^{- 1} (x + 1) - \frac{1}{2} \int \frac{2 x + 2 - 2}{x^{2} + 2 x + 2}$ $= {x t a n}^{- 1} (x + 1) - \frac{1}{2} \int \frac{d (x^{2} + 2 x + 2)}{x^{2} + 2 x + 2} + \int \frac{d x}{1 + {(x + 1)}^{2}}$ $= {x t a n}^{- 1} (x + 1) - \frac{1}{2} l n (x^{2} + 2 x + 2) + {t a n}^{- 1} (x + 1)$ $s o ∣ {x t a n}^{- 1} (x + 1) - \frac{1}{2} l n (x^{2} + 2 x + 2) + {t a n}^{- 1} (x + 1) ∣_{0}^{1}$ $= ◼ {t a n}^{- 1} (2) - \frac{1}{2} l n (5) + {t a n}^{-} (2) + \frac{1}{2} l n 2 - \frac{π}{4} ◼$ $\int_{0}^{1} {t a n}^{- 1} x d x$ $\int {t a n}^{- 1} x d x$ $= {x t a n}^{- 1} x - \int \frac{x}{1 + x^{2}} d x$ $= {x t a n}^{- 1} x - \frac{1}{2} l n (1 + x^{2})$ $s o ∣ {x t a n}^{- 1} x - \frac{1}{2} l n (1 + x^{2}) ∣_{0}^{1}$ $= {t a n}^{- 1} (1) - \frac{1}{2} l n 2$ $= \frac{π}{4} - \frac{1}{2} l n 2$ $s o a n s w e r i s$ $\int_{0}^{1} \frac{π}{2} d x + \int_{0}^{1} {t a n}^{- 1} x d x - \int_{0}^{1} {t a n}^{- 1} (x + 1) d x$ $= \frac{π}{2} + \frac{π}{4} - \frac{1}{2} l n 2 - 2 {t a n}^{- 1} (2) + \frac{1}{2} l n 5 - \frac{1}{2} l n 2 + \frac{π}{4}$ $= π - l n 2 + \frac{1}{2} l n 5 - 2 {t a n}^{- 1} (2)$ $p l s g o t h r o u g h . . .$
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