$$\text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : }$$

For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$ , where $$e=\sum_\limits{n=0}^{\infty} \frac{1}{n !}$$ and $$\mathrm{C}$$ is constant of integration, then $$\alpha+2 \beta+3 \gamma-4 \delta$$ is equal to :

If $$I(x) = \int {{e^{{{\sin }^2}x}}(\cos x\sin 2x - \sin x)dx} $$ and $$I(0) = 1$$, then $$I\left( {{\pi \over 3}} \right)$$ is equal to :

The integral $$ \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) d x $$ is equal to :