Question

Найдите неопределенный интеграл:
∫ (x-1)dx/√(5-4х-х^2)

Answer 1

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