Помогите, пожалуйста, найти длину дуги кривой y = 1/2 * x^2, 0 <= x <= 1.

y = 1/2 * x^2, 0 <= x <= 1.
L = int (0 1) (1 + ((1/2 * x^2)')^2)^(1/2) dx =
= int (0 1) (1 + (1/2 * 2 * x)^2)^(1/2) dx = int (0 1) (1 + x^2)^(1/2) dx
Пусть
I = int (1 + x^2)^(1/2) dx = x * (1 + x^2)^(1/2) - int x d((1 + x^2)^(1/2)) =
= x * (1 + x^2)^(1/2) - int x * ((1 + x^2)^(1/2))' dx =
= x * (1 + x^2)^(1/2) - int x * 1/2 * (1 + x^2)^(-1/2) * (1 + x^2)' dx =
= x * (1 + x^2)^(1/2) - int x * 1/2 * (1 + x^2)^(-1/2) * 2 * x dx =
= x * (1 + x^2)^(1/2) - int x^2/(1 + x^2)^(1/2) dx =
= x * (1 + x^2)^(1/2) - int (1 + x^2 - 1)/(1 + x^2)^(1/2) dx =
= x * (1 + x^2)^(1/2) - int (1 + x^2)^(1/2) + int dx/(1 + x^2)^(1/2) =
= x * (1 + x^2)^(1/2) - int (1 + x^2)^(1/2) + ln |x + (x^2 + 1)^(1/2)| + C
Получаем, что
I = x * (1 + x^2)^(1/2) - I + ln |x + (x^2 + 1)^(1/2)| + C
2 * I = x * (1 + x^2)^(1/2) + ln |x + (x^2 + 1)^(1/2)| + C
I = 1/2 * x * (x^2 + 1)^(1/2) + 1/2 * ln |x + (x^2 + 1)^(1/2)| + C
int (1 + x^2)^(1/2) dx = 1/2 * x * (x^2 + 1)^(1/2) + 1/2 * ln |x + (x^2 + 1)^(1/2)| + C
Получаем, что
L = int (0 1) (1 + x^2)^(1/2) dx =
= (1/2 * x * (x^2 + 1)^(1/2) + 1/2 * ln |x + (x^2 + 1)^(1/2)|)_{0}^{1} =
= (1/2 * 1 * (1^2 + 1)^(1/2) + 1/2 * ln |1 + (1^2 + 1)^(1/2)|) -
- (1/2 * 0 * (0^2 + 1)^(1/2) + 1/2 * ln |0 + (0^2 + 1)^(1/2)|) =
= 1/2 * 2^(1/2) + 1/2 * ln |1 + 2^(1/2)| = 1/2^(1/2) + 1/2 * ln (1 + 2^(1/2))
Ответ: L = 1/2^(1/2) + 1/2 * ln (1 + 2^(1/2)).

Спасибо большое!!