$\int_{0}^{1} x(1-x)^{5} dx =$

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$,where $[x]$ represents the greatest integer function,equals:

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{[x]+4} dx$, where $[\bullet]$ denotes the greatest integer function, is

$\int_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x=$

If $I_n = \int_{0}^{\pi} \frac{\sin(nx)}{\sin(x)} dx$,then the value of $\sum_{n=1}^{10} I_n$ is-

$\int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} d x=\ldots$