The solution for $x$ of the equation $\int_{\sqrt{2}}^{x} \frac{dt}{t\sqrt{t^2-1}} = \frac{\pi}{2}$ is

Let $l = \mathop {Lim}\limits_{x \to \infty } \int\limits_x^{2x} \frac{dt}{t}$ and $m = \mathop {Lim}\limits_{x \to \infty } \frac{1}{x \ln x} \int\limits_1^x \ln t \, dt$. Then the correct statement is:

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

Let $f:(0, \infty) \rightarrow \mathbb{R}$ and $F(x)=\int_0^x t f(t) d t$. If $F(x^2)=x^4+x^5$,then $\sum_{r=1}^{12} f(r^2)$ is equal to :

$\int_0^1 x \left|x - \frac{1}{2}\right| dx = $

The value of $\int_0^1 {x^2 e^x dx}$ is equal to