Задача 45861 ...

nreofn

2020-04-03 06:10:18

Найти интегралы:
∫ (x-1/sqrt(2x-1))*dx

sova

2020-04-03 09:57:43

★

Замена переменной:
sqrt(2x-1)=t
2x-1=t^2
x=(1/2)*(t^2+1);
dx=(1/2)*(t^2+1)`dt

dx=t*dt
х-1=(1/2)*(t^2+1) - 1=(1/2)(t^2+1-2)=(1/2)*(t^2-1)

∫ (x–1)dx/√2x–1)= (1/2) ∫(t^2-1)*tdt/t=(1/2) ∫ (t^2-1)dt=(1/2)*(t^3/3)-(1/2)*t+C=

=[b](1/6)*sqrt((2x-1)^3) -(1/2)*sqrt(2x-1) + C[/b]