Justify the existence of the integral $\int_1^{+\infty} \frac{\mathrm{d}t}{t\sqrt{t^2-1}}$ and show that its value is $\frac{\pi}{2}$.

To each function $f \in E$, we associate the function $U ( f )$ with derivative $U(f)^\prime(x) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Show that for all $f \in E$ and for all $x > 0$, $$\left| U ( f ) ^ { \prime } ( x ) \right| \leqslant \mathrm { e } ^ { x } \| f \| \left( \int _ { x } ^ { + \infty } \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t \right) ^ { 1 / 2 } \leqslant \| f \| \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } }$$

Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$. Show that: $$\int_0^1 h(t)\, dt = \pi.$$

The solution for $x$ of the equation $\int _ { \sqrt { 2 } } ^ { x } \frac { d t } { t \sqrt { t ^ { 2 } - 1 } } = \frac { \pi } { 2 }$ is
(1) 2
(2) $\pi$
(3) $\frac { \sqrt { 3 } } { 2 }$
(4) None of these

The value of the integral $\int _ { 1 } ^ { 2 } \left( \frac { t ^ { 4 } + 1 } { t ^ { 6 } + 1 } \right) d t$ is : (1) $\tan ^ { - 1 } \frac { 1 } { 2 } + \frac { 1 } { 3 } \tan ^ { - 1 } 8 - \frac { \pi } { 3 }$ (2) $\tan ^ { - 1 } 2 - \frac { 1 } { 3 } \tan ^ { - 1 } 8 + \frac { \pi } { 3 }$ (3) $\tan ^ { - 1 } 2 + \frac { 1 } { 3 } \tan ^ { - 1 } 8 - \frac { \pi } { 3 }$ (4) $\tan ^ { - 1 } \frac { 1 } { 2 } - \frac { 1 } { 3 } \tan ^ { - 1 } 8 + \frac { \pi } { 3 }$

The value of the integral $\int _ { 1 / 2 } ^ { 2 } \frac { \tan ^ { - 1 } x } { x } d x$ is equal to (1) $\pi \log _ { e } 2$ (2) $\frac { 1 } { 2 } \log _ { e } 2$ (3) $\frac { \pi } { 4 } \log _ { \mathrm { e } } 2$ (4) $\frac { \pi } { 2 } \log _ { \mathrm { e } } 2$