Помогите, пожалуйста, найти интеграл
int dx/(x + x^3)

int dx/(x + x^3) = int dx/(x * (1 + x^2))
Разложим подынтегральное выражение на простейшие дроби
1/(x * (1 + x^2)) = A/x + (Bx + C)/(1 + x^2) |* x * (1 + x^2)
1 = A * (1 + x^2) + (Bx + C) * x
x = 0 => 1 = A * (1 + 0^2) + (B * 0 + C) * 0 => 1 = A
x = 1 => 1 = A * (1 + 1^2) + (B * 1 + C) * 1 => 1 = 2 * A + B + C =>
=> B + C = 1 - 2 * A = 1 - 2 * 1 = -1 => B + C = -1
x = -1 => 1 = A * (1 + (-1)^2) + (B * (-1) + C) * (-1) => 1 = 2 * A + B - C =>
=> B - C = 1 - 2 * A = 1 - 2 * 1 = -1 => B - C = -1
B + C = -1, B - C = -1 => B = -1, C = 0.
Получаем, что
1/(x * (1 + x^2)) = 1/x - x/(1 + x^2)
Тогда
int dx/(x * (1 + x^2)) = int (1/x - x/(1 + x^2)) dx =
= int dx/x - int x/(1 + x^2) dx = ln |x| - int 1/(1 + x^2) d(1/2 * x^2) =
= ln |x| - 1/2 * int 1/(1 + x^2) d(x^2) = | t = x^2 | =
= ln |x| - 1/2 * int 1/(1 + t) dt = ln |x| - 1/2 * ln |1 + t| + C = | t = x^2 | =
= ln |x| - 1/2 * ln (1 + x^2) + C

Спасибо!!!