Помогите, пожалуйста, найти интеграл
int x dx/(x^2 + x + 1)
Заранее спасибо!!

int x dx/(x^2 + x + 1) = int x dx/(x^2 + 2 * x * 1/2 + 1/4 + 3/4) =
= int x dx/((x + 1/2)^2 + 3/4) = | x = -1/2 + 3^(1/2)/2 * t,
t = (x + 1/2)/(3^(1/2)/2) = (2 * x + 1)/3^(1/2), dx = 3^(1/2)/2 dt | =
= int (-1/2 + 3^(1/2)/2 * t) * 3^(1/2)/2 dt/(3/4 * t^2 + 3/4) =
= int (-3^(1/2)/4 + 3/4 * t) dt/((3/4 * (t^2 + 1)) =
= int (-3^(1/2)/4 + 3/4 * t) * 4/3 dt/(t^2 + 1) =
= int (-1/3^(1/2) + t)/(t^2 + 1) dt = -1/3^(1/2) * int dt/(t^2 + 1) + int t/(t^2 + 1) dt =
= -1/3^(1/2) * arctg t + int d(1/2 * t^2)/(t^2 + 1) =
= -1/3^(1/2) * arctg t + 1/2 * int d(t^2)/(t^2 + 1) = | t^2 = u | =
= -1/3^(1/2) * arctg t + 1/2 * int du/(u + 1) =
= -1/3^(1/2) * arctg t + 1/2 * ln |u + 1| + C = | u = t^2 | =
= -1/3^(1/2) * arctg t + 1/2 * ln (t^2 + 1) + C =
= | t = (2 * x + 1)/3^(1/2) | =
= -1/3^(1/2) * arctg ((2 * x + 1)/3^(1/2)) + 1/2 * ln (((2 * x + 1)/3^(1/2))^2 + 1) + C =
= -1/3^(1/2) * arctg ((2 * x + 1)/3^(1/2)) + 1/2 * ln ((4 * x^2 + 4 * x + 1)/3 + 1) + C =
= -1/3^(1/2) * arctg ((2 * x + 1)/3^(1/2)) + 1/2 * ln ((4 * x^2 + 4 * x + 4)/3) + C =
= -1/3^(1/2) * arctg ((2 * x + 1)/3^(1/2)) + 1/2 * ln ((4/3) * (x^2 + x + 1)) + C =
= -1/3^(1/2) * arctg ((2 * x + 1)/3^(1/2)) + 1/2 * ln (x^2 + x + 1) + C

Спасибо (IMG:style_emoticons/default/smile.gif)