Ответ:
√(5-4х-х^2) + arcsin(x/3 - 2/3) + C
Пошаговое объяснение:
∫(x-1)dx/√(5-4х-х^2) = ∫(x-1)dx/√(9-(x-2)^2)
u = x-2; du = dx
∫(u+1)du/√(9-u^2) = ∫udu/√(9-u^2) + ∫du/√(3^2-u^2)
∫udu/√(9-u^2)
v = 9-u^2; dv = -2udu; dv/-2 = udu
∫dv/(-2√v) = -√v = -√(9-u^2) = -√(9-(x-2)^2) = -√(5-4х-х^2)
∫du/√(3^2-u^2) = arcsin(u/3) = arcsin(u/3) = arcsin(x/3 - 2/3)
∫(x-1)dx/√(5-4х-х^2) = -√(5-4х-х^2) + arcsin(x/3 - 2/3) + C