Функция F(x) является первообразной функции f(x) =(10−x^2)·lnx, график которой проходит...

du = \frac{dx}{x}; (10-x^2)dx = dv => v = 10x - \frac{1}{3}x^3 | = (10x - \frac{1}{3}x^3)*lnx - \int (10x - \frac{1}{3}x^3) \frac{dx}{x} = -\frac{x}{3}(x^2-30)lnx + \frac{1}{3} \int(x^2 - 30)dx = \frac{1}{9}x^3 - 10x -\frac{x}{3}(x^2-30)lnx + c\\F(4) = 7 => \frac{1}{9}*4^3 - 10*4 -\frac{4}{3}(4^2-30)ln4 + c = 7 => c = \frac{1}{9}(359-168ln2)\\" alt="f(x) = (10-x^2)lnx\\F(x) = \int f(x)dx = \int ((10-x^2)lnx) dx = | u = lnx => du = \frac{dx}{x}; (10-x^2)dx = dv => v = 10x - \frac{1}{3}x^3 | = (10x - \frac{1}{3}x^3)*lnx - \int (10x - \frac{1}{3}x^3) \frac{dx}{x} = -\frac{x}{3}(x^2-30)lnx + \frac{1}{3} \int(x^2 - 30)dx = \frac{1}{9}x^3 - 10x -\frac{x}{3}(x^2-30)lnx + c\\F(4) = 7 => \frac{1}{9}*4^3 - 10*4 -\frac{4}{3}(4^2-30)ln4 + c = 7 => c = \frac{1}{9}(359-168ln2)\\" align="absmiddle" class="latex-formula">

$F(x) = \frac{1}{9}x^3 - 10x -\frac{x}{3}(x^2-30)lnx + \frac{1}{9}(359-168ln2)\\\\F(2) = \frac{1}{9}(187-12ln2)\\ F(3) = -27+\frac{1}{9}(359-168ln2)+21ln3$

Вроде бы нет никакого соотношения)