помогите решить)
1) int arctg (2x - 1)^(1/2) dx
2) int (x^2 - 3x) * sin 2x dx
Заранее спасибо!!

1) int arctg (2x - 1)^(1/2) dx = x * arctg (2x - 1)^(1/2) - int x d(arctg (2x - 1)^(1/2)) =
= x * arctg (2x - 1)^(1/2) - int x * 1/(1 + ((2x - 1)^(1/2))^2) * 1/(2x - 1)^(1/2) dx =
= x * arctg (2x - 1)^(1/2) - int x * 1/(1 + 2x - 1) * 1/(2x - 1)^(1/2) dx =
= x * arctg (2x - 1)^(1/2) - int x * 1/(2x) * 1/(2x - 1)^(1/2) dx =
= x * arctg (2x - 1)^(1/2) - 1/2 * int dx/(2x - 1)^(1/2) =
= x * arctg (2x - 1)^(1/2) - 1/2 * int (2x - 1)^(-1/2) dx =
= x * arctg (2x - 1)^(1/2) - 1/2 * 1/2 * 1/(-1/2 + 1) * (2x - 1)^(-1/2 + 1) + C =
= x * arctg (2x - 1)^(1/2) - 1/2 * (2x - 1)^(1/2) + C
2) int (x^2 - 3x) * sin 2x dx = int (x^2 - 3x) d(-1/2 * cos 2x) =
= -1/2 * int (x^2 - 3x) d(cos 2x) =
= -1/2 * (x^2 - 3x) * cos 2x + 1/2 * int cos 2x d(x^2 - 3x) =
= -1/2 * (x^2 - 3x) * cos 2x + 1/2 * int (2x - 3) * cos 2x dx =
= -1/2 * (x^2 - 3x) * cos 2x + 1/2 * int (2x - 3) d(1/2 * sin 2x) =
= -1/2 * (x^2 - 3x) * cos 2x + 1/4 * int (2x - 3) d(sin 2x) =
= -1/2 * (x^2 - 3x) * cos 2x + 1/4 * (2x - 3) * sin 2x - 1/4 * int sin 2x d(2x - 3) =
= -1/2 * (x^2 - 3x) * cos 2x + 1/4 * (2x - 3) * sin 2x - 1/2 * int sin 2x dx =
= -1/2 * (x^2 - 3x) * cos 2x + 1/4 * (2x - 3) * sin 2x + 1/2 * 1/2 * cos 2x + C =
= -1/2 * (x^2 - 3x) * cos 2x + 1/4 * (2x - 3) * sin 2x + 1/4 * cos 2x + C =
= -1/2 * (x^2 - 3x - 1/2) * cos 2x + 1/4 * (2x - 3) * sin 2x + C