Помогите, пожалуйста, найти интегралы:
1) int x^3/(1 - x^4)^(1/2) dx
2) int (x^2 + 3)/(x^2 + 3 * x + 2) dx

1) int x^3/(1 - x^4)^(1/2) dx = int 1/(1 - x^4)^(1/2) d(1/4 * x^4) =
= 1/4 * int 1/(1 - x^4)^(1/2) d(x^4) = | t = x^4 | =
= 1/4 * int 1/(1 - t)^(1/2) dt = 1/4 * int (1 - t)^(-1/2) dt =
= 1/4 * 1/(-1) * 1/(-1/2 + 1) * (1 - t)^(-1/2 + 1) + C =
= -1/2 * (1 - t)^(1/2) + C = | t = x^4 | = -1/2 * (1 - x^4)^(1/2) + C
2) int (x^2 + 3)/(x^2 + 3 * x + 2) dx =
= int (x^2 + 3 * x + 2 - 3 * x + 1)/(x^2 + 3 * x + 2) dx =
= int dx - int (3 * x - 1)/(x^2 + 3 * x + 2) dx =
= x - int (3 * x - 1)/((x + 1) * (x + 2)) dx
Разложим подынтегральное выражение на простейшие дроби
(3 * x - 1)/((x + 1) * (x + 2)) = A/(x + 1) + B/(x + 2) |* (x + 1) * (x + 2)
3 * x - 1 = A * (x + 2) + B * (x + 1)
x = -2 => 3 * (-2) - 1 = B * (-2 + 1) => -7 = -B => B = 7
x = -1 => 3 * (-1) - 1 = A * (-1 + 2) => A = -4.
Получаем, что
(3 * x - 1)/((x + 1) * (x + 2)) = -4/(x + 1) + 7/(x + 2), тогда
x - int (3 * x - 1)/((x + 1) * (x + 2)) dx =
= x - int (-4/(x + 1) + 7/(x + 2)) dx = x + 4 * ln |x + 1| - 7 * ln |x + 2| + C