Помогите, пожалуйста, найти интеграл
int (4 * x + 2) dx/(1 - (x + 1)^2)^(1/2) dx
int (4 * x + 2) dx/(1 - (x + 1)^2)^(1/2) dx = | t = x + 1; x = t - 1; dx = dt | =
= int (4 * (t - 1) + 2) dt/(1 - t^2)^(1/2) = int (4 * t - 2) dt/(1 - t^2)^(1/2) =
= int 4 * t dt/(1 - t^2)^(1/2) - int 2 dt/(1 - t^2)^(1/2) =
= 2 * int d(t^2)/(1 - t^2)^(1/2) - 2 * arcsin t = | t^2 = u | =
= 2 * int du/(1 - u)^(1/2) - 2 * arcsin t = 2 * int (1 - u)^(-1/2) du - 2 * arcsin t =
= 2 * (-1) * 1/(-1/2 + 1) * (1 - u)^(-1/2 + 1) - 2 * arcsin t + C =
= -4 * (1 - u)^(1/2) - 2 * arcsin t + C = | u = t^2 | =
= -4 * (1 - t^2)^(1/2) - 2 * arcsin t + C = | t = x + 1 | =
= -4 * (1 - (x + 1)^2)^(1/2) - 2 * arcsin (x + 1) + C
Спасибо!!
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