МаксиM
Сообщение
#11836 1.3.2008, 16:22
Помогите, пожалуйста, найти интеграл
int x^2/(x^3 + 1)^(1/6) dx
tig81
Сообщение
#11838 1.3.2008, 16:27
int x^2/(x^3 + 1)^(1/6) dx = int 1/(x^3 + 1)^(1/6) d(1/3 * x^3) =
= 1/3 * int 1/(x^3 + 1)^(1/6) d(x^3) = | x^3 = t | =
= 1/3 * int 1/(t + 1)^(1/6) dt = 1/3 * int (t + 1)^(-1/6) dt =
= 1/3 * 1/(-1/6 + 1) * (t + 1)^(-1/6 + 1) + C =
= 1/3 * 6/5 * (t + 1)^(5/6) + C = 2/5 * (t + 1)^(5/6) + C = | t = x^3 | =
= 2/5 * (x^3 + 1)^(5/6) + C