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Question Number 119006 by A8;15: last updated on 21/Oct/20

	Answered by Dwaipayan Shikari last updated on 21/Oct/20

	$\int_{\frac{1}{2}}^{2} \frac{{t a n}^{- 1} x}{x^{2} - x + 1} d x = I$ $= \int_{2}^{\frac{1}{2}} \frac{{t a n}^{- 1} \frac{1}{t}}{\frac{1}{t^{2}} - \frac{1}{t} + 1} . \frac{d t}{- t^{2}} x = \frac{1}{t} \Rightarrow 1 = \frac{1}{- t^{2}} . \frac{d t}{d x}$ $= \int_{\frac{1}{2}}^{2} \frac{\frac{π}{2} - {t a n}^{- 1} t}{t^{2} - t + 1} d t = I$ $2 I = \int_{\frac{1}{2}}^{2} \frac{\frac{π}{2}}{t^{2} - t + 1} d t$ $4 I = π \int_{\frac{1}{2}}^{2} \frac{1}{t^{2} - t + 1} d t$ $\frac{4 I}{π} = \int_{\frac{1}{2}}^{2} \frac{1}{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d t$ $\frac{4 I}{π} = \frac{2}{\sqrt{3}} {[{t a n}^{- 1} \frac{2 t - 1}{\sqrt{3}}]}_{\frac{1}{2}}^{2}$ $I = \frac{π}{4} . \frac{2}{\sqrt{3}} . \frac{π}{3} = \frac{π^{2}}{6 \sqrt{3}}$
	Commented by A8;15: last updated on 21/Oct/20
	thanks sir
	Answered by Bird last updated on 22/Oct/20
	$we put for a>0 f(a)=∫_(1/a) ^a ((arctanx)/(x^2 −x+1))dx ⇒f(a)=_(x=(1/t)) −∫_(1/a) ^a (((π/2)−arctant)/((1/t^2 )−(1/t)+1))(−(dt/t^2 )) =∫_(1/a) ^a (((π/2)−arctant)/(1−t +t^2 ))dt =(π/2)∫_(1/a) ^a (dt/(t^2 −t+1))−f(a) ⇒ 2f(a) =(π/2)∫_(1/a) ^a (dt/((t−(1/2))^2 +(3/4))) =_(t−(1/2)=((√3)/2)z) (π/2)×(4/3)∫_(((2/a)−1)/( (√3))) ^((2a−1)/( (√3))) (1/(1+z^2 ))×((√3)/2)dz =((2π)/3)×((√3)/2){arctan(((2a−1)/( (√3))))−arctan((((2/a)−1)/( (√3))))} =(π/( (√3))){arctan(((2a−1)/( (√3))))−arctan((((2/a)−1)/( (√3))))} 2f(2)=(π/( (√3))){ arctan((√3))−arctan(0)} =(π/( (√3)))×(π/3) =(π^2 /(3(√3))) ⇒f(2)=(π^2 /(6(√3)))$
	$w e p u t f o r a > 0 f (a) = \int_{\frac{1}{a}}^{a} \frac{a r c t a n x}{x^{2} - x + 1} d x$ $\Rightarrow f (a) =_{x = \frac{1}{t}} - \int_{\frac{1}{a}}^{a} \frac{\frac{π}{2} - a r c t a n t}{\frac{1}{t^{2}} - \frac{1}{t} + 1} (- \frac{d t}{t^{2}})$ $= \int_{\frac{1}{a}}^{a} \frac{\frac{π}{2} - a r c t a n t}{1 - t + t^{2}} d t$ $= \frac{π}{2} \int_{\frac{1}{a}}^{a} \frac{d t}{t^{2} - t + 1} - f (a) \Rightarrow$ $2 f (a) = \frac{π}{2} \int_{\frac{1}{a}}^{a} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}}$ $=_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} z} \frac{π}{2} \times \frac{4}{3} \int_{\frac{\frac{2}{a} - 1}{\sqrt{3}}}^{\frac{2 a - 1}{\sqrt{3}}} \frac{1}{1 + z^{2}} \times \frac{\sqrt{3}}{2} d z$ $= \frac{2 π}{3} \times \frac{\sqrt{3}}{2} {a r c t a n (\frac{2 a - 1}{\sqrt{3}}) - a r c t a n (\frac{\frac{2}{a} - 1}{\sqrt{3}})}$ $= \frac{π}{\sqrt{3}} {a r c t a n (\frac{2 a - 1}{\sqrt{3}}) - a r c t a n (\frac{\frac{2}{a} - 1}{\sqrt{3}})}$ $2 f (2) = \frac{π}{\sqrt{3}} {a r c t a n (\sqrt{3}) - a r c t a n (0)}$ $= \frac{π}{\sqrt{3}} \times \frac{π}{3} = \frac{π^{2}}{3 \sqrt{3}} \Rightarrow f (2) = \frac{π^{2}}{6 \sqrt{3}}$
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