Question

已知导数求原函数!急!f'(x)=√(1-x^2)怎么求f(x)?

Answer 1

∫√(1-x^2)dx=x√(1-x^2)+∫x^2/√(1-x^2)dx=x√(1-x^2)+∫[(-√(1-x^2)+1/√(1-x^2)]dx

故∫√(1-x^2=【x√(1-x^2)+∫1/√(1-x^2)dx】/2=【x√(1-x^2)+acsinx】/2