For real numbers $\alpha ,\beta ,\gamma$ and $\delta$ , if $\int \frac{\left(x^2-1\right)+{\tan }^{-1}\left(\frac{x^2+1}{x}\right)}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}dx=\alpha {\log }_e\left({\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right)+\beta {\tan }^{-1}\left(\frac{\gamma \left(x^2-1\right)}{x}\right)+\delta {\tan }^{-1}\left(\frac{x^2+1}{x}\right)+C$

where C is an arbitrary constant, then the value of $10\left(\alpha +\beta \gamma +\delta \right)$ is equal to _____

$\begin{array}{l}G.E.=I=\int \frac{\left(x^2-1\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}+\int \frac{dx}{\left(x^4+3x^2+1\right)}I=\int \frac{x^2\left(1-\frac{1}{x^2}\right)dx}{x^2\left(x^2+3+\frac{1}{x^2}\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}+\int \frac{dx}{x^2\left(x^2+3+\frac{1}{x^2}\right)},\\ Now, I_1=\int \frac{\left(1-\frac{1}{x^2}\right)dx}{\left({\left(x+\frac{1}{x}\right)}^2+1\right){\tan }^{-1}\left(x+\frac{1}{x}\right)},\left(\text{Take }{\tan }^{-1}\left(x+\frac{1}{x}\right)=\theta , x+\frac{1}{x}=\tan \theta , \left(1-\frac{1}{x^2}\right) dx={\sec }^2\theta d\theta \right)\end{array}$

$\begin{array}{l} =\int \frac{{\sec }^2\theta d\theta }{\left(1+{\tan }^2\theta \right)\theta }=\log \left|\theta \right|=\log \left|{\tan }^{-1}\left(x+\frac{1}{x}\right)\right| \\ and I_2=\frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+3+\frac{1}{x^2}}dx\\ =\frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)dx}{{\left(x-\frac{1}{x}\right)}^2+5}-\frac{1}{2}\int \frac{\left(1-\frac{1}{x^2}\right)dx}{{\left(x+\frac{1}{x}\right)}^2+1}\\ =\frac{1}{2}\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{5}}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{x+\frac{1}{x}}{1}\right)\end{array}$

$\begin{array}{l}\therefore I=\log \left({\tan }^{-1}\left(x+\frac{1}{x}\right)\right)+\frac{1}{2}\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{5}}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{x+\frac{1}{x}}{1}\right)\\ =\log \left({\tan }^{-1}\left(x+\frac{1}{x}\right)\right)+\frac{1}{2}\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{\frac{1}{\sqrt{5}}\left(x^2-1\right)}{x}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{x^2+1}{x}\right)\\ Comparing, we get \alpha =1,\beta =\frac{1}{2\sqrt{5}},\gamma =\frac{1}{\sqrt{5}},\delta =\frac{-1}{2} \\ \therefore 10(\alpha +\beta \gamma +\delta )=10\left(1+\frac{1}{10}-\frac{1}{2}\right)=10+1-5=6\end{array}$