The function $L(x) = \int_1^x {\frac{{dt}}{t}} $ satisfies the equation

Number of values of $x$ satisfying the equation

$\int\limits_{ - \,1}^x {\,\left( {8{t^2} + \frac{{28}}{3}t + 4} \right)\,dt} $ $=$ $\frac{{\left( {{\textstyle{3 \over 2}}} \right)x + 1}}{{{{\log }_{(x + 1)}}\sqrt {x + 1} }}$ , is

Let $J=\int_0^1 \frac{x}{1+x^8} d x$

Consider the following assertions:

$I$. $J>\frac{1}{4}$

$II$. $J<\frac{\pi}{8}$ Then,

If ${I_1} = \int_0^1 {{2^{{x^2}}}dx,\;} {I_2} = \int_0^1 {{2^{{x^3}}}dx} ,\;{I_3} = \int_1^2 {{2^{{x^2}}}dx} $,${I_4} = \int_1^2 {{2^{{x^3}}}dx} $, then

Let $f: R \rightarrow R$ be a function defined as $f(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in R$, where [t] is the greatest integer less than or equal to $t$. If $\lim _{x \rightarrow-1} f(x)$ exists, then the value of $\int_{0}^{4} f(x) d x$ is equal to.

The true solution set of the inequality,$\sqrt{5x-6-x^2}+\left( \frac{\pi }{2}\int\limits_{0}^{x}{dz} \right)>x\int\limits_{0}^{\pi }{{{\sin }^{2}}xdx}$ is: