find-the-value-of-0-arctanx-x-2-x-1-dx-

Question Number 31967 by abdo imad last updated on 17/Mar/18

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{a r c t a n x}{x^{2} + x + 1} d x .

Commented by abdo imad last updated on 19/Mar/18

l e t p u t I = \int_{0}^{\infty} \frac{a r c t a n x}{x^{2} + x + 1} d x . c h . x = \frac{1}{t} g i v e

I = \int_{0}^{\infty} \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_{0}^{\infty} \frac{\frac{π}{2} - a r c t a n t}{1 + t + t^{2}} d t

= \frac{π}{2} \int_{0}^{\infty} \frac{d t}{t^{2} + t + 1} - I \Rightarrow 2 I = \frac{π}{2} \int_{0}^{\infty} \frac{d t}{t^{2} + t + 1} \Rightarrow

I = \frac{π}{4} \int_{0}^{\infty} \frac{d t}{t^{2} + t + 1} . b u t

\int_{0}^{\infty} \frac{d t}{t^{2} + t + 1} = \int_{0}^{\infty} \frac{d t}{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} (. c h . t + \frac{1}{2} = \frac{\sqrt{3}}{2} u)

= \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{1}{\frac{3}{4} (t^{2} + 1)} \frac{\sqrt{3}}{2} d u = \frac{4}{3} \frac{\sqrt{3}}{2} \int_{\frac{1}{\sqrt{3}}}^{+ \infty} \frac{d u}{1 + u^{2}}

= \frac{2 \sqrt{3}}{3} {[a r c t a n (u)]}_{\frac{1}{\sqrt{3}}}^{\infty} = \frac{2 \sqrt{3}}{3} (\frac{π}{2} - a r c t a n (\frac{1}{\sqrt{3}}))

= \frac{2 \sqrt{3}}{3} a r c t a n (\sqrt{3}) = \frac{2 \sqrt{3}}{3} \frac{π}{3} = \frac{2 π \sqrt{3}}{9} \Rightarrow

I = \frac{π}{4} \frac{2 π \sqrt{3}}{9} = \frac{π^{2} \sqrt{3}}{18} .