Здравствуйте! Помогите, пожалуйста, решить (подскажите принцип) 2 интеграла:
1. int (x^2 + 2) dx/(x^3 + x^2 - 2 * x)
2. int (2 * x^5 - 3 * x^2) dx/(1 + 3 * x^3 - x^6)
Заранее благодарен.

1. int (x^2 + 2) dx/(x^3 + x^2 - 2 * x) = int (x^2 + 2) dx/(x * (x^2 + x - 2)) =
= int (x^2 + 2) dx/(x * (x - 1) * (x + 2))
Разложим подинтегральное выражение на простейшие дроби:
(x^2 + 2)/(x * (x - 1) * (x + 2)) = A/x + B/(x - 1) + C/(x + 2)
Домножим обе части равенства на x * (x - 1) * (x + 2)
x^2 + 2 = A * (x - 1) * (x + 2) + B * x * (x + 2) + C * x * (x - 1)
x = 0 => 2 = -2A => A = -1
x = 1 => 3 = 3B => B = 1
x = -2 => 6 = 6C => C = 1
Тогда
int (x^2 + 2) dx/(x * (x - 1) * (x + 2)) = int (-1/x + 1/(x - 1) + 1/(x + 2)) dx =
= -ln |x| + ln |x - 1| + ln |x + 2| + C
2. int (2 * x^5 - 3 * x^2) dx/(1 + 3 * x^3 - x^6) =
= | t = 1 + 3 * x^3 - x^6; dt = 9 * x^2 - 6 * x^5 | =
= -1/3 * int dt/t = -1/3 * ln |t| + C = | t = 1 + 3 * x^3 - x^6 | =
= -1/3 * ln |1 + 3 * x^3 - x^6| + C

Большое спасибо, разобрался.

Еще раз добрый день! Наведите, плз, на мысль еще по двум интегралам.
1. int dx/(x^2 + 2 * x + 3)
2. int (x + 2) dx/(x^2 + 2 * x + 2)
Спасибо.

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