The integral ∫01cot−1(1+x+x2)dx\int\limits_{0}^{1} \cot^{-1} (1+x+x^{2}) dx0∫1cot−1(1+x+x2)dx is equal to :
2tan−12+12loge(54)+π22 \tan^{-1} 2 + \dfrac{1}{2} \log_{e} \left( \dfrac{5}{4} \right) + \dfrac{\pi}{2}2tan−12+21loge(45)+2π
2tan−12+12loge(54)−π22 \tan^{-1} 2 + \dfrac{1}{2} \log_{e} \left( \dfrac{5}{4} \right) - \dfrac{\pi}{2}2tan−12+21loge(45)−2π
2tan−12−12loge(54)+π22 \tan^{-1} 2 - \dfrac{1}{2} \log_{e} \left( \dfrac{5}{4} \right) + \dfrac{\pi}{2}2tan−12−21loge(45)+2π
2tan−12−12loge(54)−π22 \tan^{-1} 2 - \dfrac{1}{2} \log_{e} \left( \dfrac{5}{4} \right) - \dfrac{\pi}{2}2tan−12−21loge(45)−2π