Помогите, пожалуйста, вычислить интеграл:
int (1 4) (3 * x^2 + 2 * x + 1) * ln (x^(1/2)/2) dx

Сначала вычислим неопределенный интеграл:
int (3 * x^2 + 2 * x + 1) * ln (x^(1/2)/2) dx = int ln (x^(1/2)/2) d(x^3 + x^2 + x) =
= ln (x^(1/2)/2) * (x^3 + x^2 + x) - int (x^3 + x^2 + x) d(ln (x^(1/2)/2)) =
= ln (x^(1/2)/2) * (x^3 + x^2 + x) -
- int (x^3+x^2+x) * 1/(x^(1/2)/2) * (x^(1/2)/2)' dx =
= ln (x^(1/2)/2) * (x^3 + x^2 + x) -
- int (x^3 + x^2 + x) * 1/(x^(1/2)/2) * 1/2 * 1/2 * x^(-1/2) dx =
= ln (x^(1/2)/2) * (x^3 + x^2 + x) - 1/2 * int (x^3 + x^2 + x) * 1/x dx =
= ln (x^(1/2)/2) * (x^3 + x^2 + x) - 1/2 * int (x^2 + x + 1) dx =
= ln (x^(1/2)/2) * (x^3 + x^2 + x) - 1/2 * (1/3 * x^3 + 1/2 * x^2 + x) + C =
= ln (x^(1/2)/2) * (x^3 + x^2 + x) - 1/6 * x^3 - 1/4 * x^2 - 1/2 * x + C
Получаем, что
int (1 4) (3 * x^2 + 2 * x + 1) * ln (x^(1/2)/2) dx =
= (ln (x^(1/2)/2) * (x^3 + x^2 + x) - 1/6 * x^3 - 1/4 * x^2 - 1/2 * x)_{1}^{4} =
= (ln (4^(1/2)/2) * (4^3 + 4^2 + 4) - 1/6 * 4^3 - 1/4 * 4^2 - 1/2 * 4) -
- (ln (1^(1/2)/2) * (1^3 + 1^2 + 1) - 1/6 * 1^3 - 1/4 * 1^2 - 1/2 * 1) =
= (ln 1 * (4^3 + 4^2 + 4) - 64/6 - 4 - 2) - (ln 1/2 * (1 + 1 + 1) - 1/6 - 1/4 - 1/2) =
= (-64/6 - 4 - 2) - (-ln 2 * 3 - 1/6 - 1/4 - 1/2) =
= -32/3 - 6 + 3 * ln 2 + 1/6 + 1/4 + 1/2 = 3 * ln 2 - 63/4.

Спасибо.

Пожалуйста