设∫xf(x)dx=arcsinx+c,求∫1/f(x)dx

∫xf(x)dx=arcsinx+C
xf(x) = 1/√(1-x^2)
1/f(x) = x√(1-x^2)
∫dx/f(x) =∫x√(1-x^2) dx
let
x= siny
dx = cosy dy
∫dx/f(x)
=∫x√(1-x^2) dx
=∫siny(cosy)^2 dy
=(1/3)(cosy)^3 + C'
=(1/3)(1-x^2)^(3/2) + C