已知∫xf(x)dx=arcsinx+C，求∫1⼀f(x)dx

解：∵(arcsinx)'=xf(x)=(1-x^2)^ (-1/2)
∴f(x)=[x (1-x^2)^ 1/2] ^(-1)
1/f(x)=x(1-x^2) ^1/2
∫1/f(x)dx =∫x(1-x^2) ^1/2dx
=-1/2∫(1-x^2)^ 1/2 d(1-x^2)
= -1/3(1-x^2) ^(3/2) + C

y = arcsinx
siny = x
cosy dy/dx = 1
dy/dx = 1/cosy = 1/ √(1-x^2)

∫xf(x)dx=arcsinx+C
d/dx(∫xf(x)dx) = d/dx(arcsinx+C)
=> xf(x) = 1/√(1-x^2)
1/f(x) = x √(1-x^2)
∫1/f(x)dx = ∫ x √(1-x^2) dx
= -(1/2)∫ √(1-x^2) d(1-x^2)
= -1/3(1-x^2)^(3/2) + C