Elena
Сообщение
#10065 13.1.2008, 10:50
Помогите, пожалуйста, найти интеграл
int (x^3 + 2)/(x^2 - x - 2) dx
Dimka
Сообщение
#10066 13.1.2008, 11:07
int (x^3 + 2)/(x^2 - x - 2) dx = int (x * (x^2 - x - 2) + x^2 + 2 * x + 2)/(x^2 - x - 2) dx =
= int x dx + int (x^2 + 2 * x + 2)/(x^2 - x - 2) dx =
= 1/2 * x^2 + int ((x^2 - x - 2) + (3 * x + 4))/(x^2 - x - 2) dx =
= 1/2 * x^2 + int dx + int (3 * x + 4))/(x^2 - x - 2) dx =
= 1/2 * x^2 + x + int (3 * x + 4))/((x + 1) * (x - 2)) dx =
= 1/2 * x^2 + x + int (3 * x + 3 + 1)/((x + 1) * (x - 2)) dx =
= 1/2 * x^2 + x + int (3 * x + 3)/((x + 1) * (x - 2)) dx + int dx/((x + 1) * (x - 2)) =
= 1/2 * x^2 + x + int 3/(x - 2) dx + 1/3 * int 3/((x + 1) * (x - 2)) dx =
= 1/2 * x^2 + x + 3 * ln |x - 2| + 1/3 * int ((x + 1) - (x - 2))/((x + 1) * (x - 2)) dx =
= 1/2 * x^2 + x + 3 * ln |x - 2| + 1/3 * int dx/(x - 2) - 1/3 * int dx/(x + 1) =
= 1/2 * x^2 + x + 3 * ln |x - 2| + 1/3 * ln |x - 2| - 1/3 * ln |x + 1| + C =
= 1/2 * x^2 + x + 10/3 * ln |x - 2| - 1/3 * ln |x + 1| + C