$$ \int\left[\frac{x^4-x}{x^{20}}\right]^{1 / 4} d x= $$

$$ \begin{array}{r}\mathop {\lim }\limits_{n \to \infty }\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}\right. \\ +\frac{n^{3 / 2}}{(n+4)^{5 / 2}}-\frac{n^{1 / 2}}{(n+6)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}} \\ \left.-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]= \end{array} $$

$$ \lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]= $$

If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4 x=0$, the parabola $y^2=x$ and the $X$-axis is $A$, then $6 A-9 \sqrt{3}=$