Алгебра: Решите неравенства Б и В

Б
$\int\limits^x_0 {x(3t^2-8t+3) \, dt =t^3-4t^2+3t|^x_{0}=x^3-4x^2+3x\ \textgreater \ 0$
x(x²-4x+3)>0
x(x-1)(x-3)>0
x=0 x=1 x=3
_ + _ +
------------------(0)-----------(1)----------------(3)---------------------
x∈(0;1) U (3;∞)

$\int\limits^x_0 {t^3} \, dt =t^4/4|^x_{0}=x^4/4\ \textless \ 1/4$
x^4-1<0<br>(x-1)(x+1)(t²+1)<0<br>x=1 x=-1
+ _ +
-----------------(-1)-----------------(1)-----------------
x∈(-1;1)