For real numbers α, β, γ and , if ∫ ( x ^ 2 - 1) + ^ - 1 ( x ^ 2 +...

$∫ \dfrac{\left({x}^{2} - 1\right)dx}{\left({x}^{4} + 3{x}^{2} + 1\right)\tan^{- 1}\left(x + \dfrac{1}{x}\right)} + ∫ \dfrac{dx}{{x}^{4} + 3{x}^{2} + 1}$

$∫ \dfrac{\left(1 - \dfrac{1}{{x}^{2}}\right)dx}{\left({\left(x + \dfrac{1}{x}\right)}^{2} + 1\right)\tan^{- 1}\left(x + \dfrac{1}{x}\right)} + \dfrac{1}{2} ∫ \dfrac{\left({x}^{2} + 1\right) - \left({x}^{2} - 1\right)dx}{{x}^{4} + 3{x}^{2} + 1}$

Put $\tan^{- 1}\left(x + \dfrac{1}{x}\right) = t$

$∫ \dfrac{dt}{t} + \dfrac{1}{2} ∫ \dfrac{\left(1 + \dfrac{1}{{x}^{2}}\right)dx}{{\left(x - \dfrac{1}{x}\right)}^{2} + 5} - \dfrac{1}{2} ∫ \dfrac{\left(1 - \dfrac{1}{{x}^{2}}\right)dx}{{\left(x + \dfrac{1}{x}\right)}^{2} + 1}$

Put $x - \dfrac{1}{x} = y, x + \dfrac{1}{x} = z$

$\log_{e}t + \dfrac{1}{2} ∫ \dfrac{dy}{{y}^{2} + 5} - \dfrac{1}{2} ∫ \dfrac{dz}{{z}^{2} + 1}$

$= \log_{\text{e}}\tan^{- 1}\left(x + \dfrac{1}{x}\right) + \dfrac{1}{2\sqrt{5}}\tan^{- 1}\left(\dfrac{{x}^{2} - 1}{\sqrt{5}x}\right)$

$- \dfrac{1}{2}\tan^{- 1}\left(\dfrac{{x}^{2} + 1}{x}\right) + C$

$\alpha = 1, \beta = \dfrac{1}{2\sqrt{5}}, \gamma = \dfrac{1}{\sqrt{5}}, \delta = \dfrac{- 1}{2}$

or $\alpha = 1, \beta = \dfrac{- 1}{2\sqrt{5}}, \gamma = \dfrac{- 1}{\sqrt{5}}, \delta = \dfrac{- 1}{2}$

$10 ( \alpha + \beta\gamma + \delta ) = 10\left(1 + \dfrac{1}{10} - \dfrac{1}{2}\right) = 6$