For real numbers $$\alpha, \beta, \gamma$$ and $$\delta$$, if $$\int \frac{(x^2-1)+\tan^{-1}\left(\frac{x^2+1}{x}\right)}{(x^4+3x^2+1)\tan^{-1}\left(\frac{x^2+1}{x}\right)}dx = \alpha\log_e\left(\tan^{-1}\left(\frac{x^2+1}{x}\right)\right) + \beta\tan^{-1}\left(\frac{\gamma(x^2-1)}{x}\right) + \delta\tan^{-1}\left(\frac{x^2+1}{x}\right) + C$$ where $$C$$ is an arbitrary constant, then the value of $$10(\alpha + \beta\gamma + \delta)$$ is equal to ________.

We need to evaluate $$\int \frac{(x^2-1)+\tan^{-1}\left(\frac{x^2+1}{x}\right)}{(x^4+3x^2+1)\tan^{-1}\left(\frac{x^2+1}{x}\right)}dx$$.

Let $$t = \tan^{-1}\left(\frac{x^2+1}{x}\right) = \tan^{-1}\left(x+\frac{1}{x}\right)$$. The derivative of $$x + \frac{1}{x}$$ is $$1 - \frac{1}{x^2} = \frac{x^2-1}{x^2}$$, and $$1 + \left(x+\frac{1}{x}\right)^2 = x^2 + 3 + \frac{1}{x^2} = \frac{x^4+3x^2+1}{x^2}$$. Therefore $$\frac{dt}{dx} = \frac{x^2-1}{x^4+3x^2+1}$$.

Splitting the integrand: $$\int \frac{x^2-1}{(x^4+3x^2+1)\,t}\,dx + \int \frac{1}{x^4+3x^2+1}\,dx = \int \frac{dt}{t} + \int \frac{dx}{x^4+3x^2+1}$$.

The first integral gives $$\ln|t| = \log_e\left|\tan^{-1}\left(\frac{x^2+1}{x}\right)\right|$$.

For the second integral, we divide numerator and denominator by $$x^2$$ to get $$\int \frac{1/x^2}{x^2+3+1/x^2}\,dx$$. We decompose $$\frac{1}{x^2} = \frac{1}{2}\left(1+\frac{1}{x^2}\right) - \frac{1}{2}\left(1-\frac{1}{x^2}\right)$$.

For the first part, substituting $$u = x - \frac{1}{x}$$ so that $$du = \left(1+\frac{1}{x^2}\right)dx$$ and $$x^2+3+\frac{1}{x^2} = u^2 + 5$$: $$\frac{1}{2}\int \frac{du}{u^2+5} = \frac{1}{2} \cdot \frac{1}{\sqrt{5}}\tan^{-1}\frac{u}{\sqrt{5}} = \frac{1}{2\sqrt{5}}\tan^{-1}\frac{x^2-1}{\sqrt{5}\,x}$$.

For the second part, substituting $$v = x + \frac{1}{x}$$ so that $$dv = \left(1-\frac{1}{x^2}\right)dx$$ and $$x^2+3+\frac{1}{x^2} = v^2 + 1$$: $$-\frac{1}{2}\int \frac{dv}{v^2+1} = -\frac{1}{2}\tan^{-1}(v) = -\frac{1}{2}\tan^{-1}\left(\frac{x^2+1}{x}\right)$$.

Combining everything: $$\log_e\left(\tan^{-1}\frac{x^2+1}{x}\right) + \frac{1}{2\sqrt{5}}\tan^{-1}\frac{x^2-1}{\sqrt{5}\,x} - \frac{1}{2}\tan^{-1}\frac{x^2+1}{x} + C$$.

Matching with the given form $$\alpha\log_e\left(\tan^{-1}\frac{x^2+1}{x}\right) + \beta\tan^{-1}\left(\frac{\gamma(x^2-1)}{x}\right) + \delta\tan^{-1}\frac{x^2+1}{x} + C$$, we identify $$\alpha = 1$$, $$\gamma = \frac{1}{\sqrt{5}}$$, $$\beta = \frac{1}{2\sqrt{5}}$$, and $$\delta = -\frac{1}{2}$$.

Therefore $$10(\alpha + \beta\gamma + \delta) = 10\left(1 + \frac{1}{2\sqrt{5}} \cdot \frac{1}{\sqrt{5}} - \frac{1}{2}\right) = 10\left(1 + \frac{1}{10} - \frac{1}{2}\right) = 10 \times \frac{3}{5} = 6$$.

The answer is $$6$$.