Есть интеграл int (x + x^(2/3) + x^(1/6))/(x * (1 + x^(1/3)) dx
Решаю через замену x^(1/6) = t
получаю 6 * int (t^5 + t^3 + 1)/(1+t^2)dt, а дальше надо разбивать на сумму 3 интегралов?
Подскажите пожалуйста. Заранее благодарен.
int (x + x^(2/3) + x^(1/6))/(x * (1 + x^(1/3)) dx =
= | x^(1/6) = t; x = t^6; dx = 6 * t^5 dt | =
= int (t^6 + t^4 + t)/(t^6 * (1 + t^2)) * 6 * t^5 dt =
= 6 * int (t^6 + t^4 + t)/(t * (1 + t^2)) dt = 6 * int (t^5 + t^3 + 1)/(1 + t^2) dt =
= 6 * int (t^5 + t^3)/(t^2 + 1) dt + 6 * int dt/(1 + t^2) =
= 6 * int t^3 * (t^2 + 1)/(t^2 + 1) dt + 6 * int dt/(1 + t^2) =
= 6 * int t^3 dt + 6 * arctg t = 6 * 1/4 * t^4 + 6 * arctg t + C =
= | t = x^(1/6) | = 3/2 * x^(2/3) + 6 * arctg x^(1/6) + C
Спасибо!!!
Русская версия Invision Power Board (http://www.invisionboard.com)
© Invision Power Services (http://www.invisionpower.com)