高数,积分.设f（x）dx为x^2/(1+x)^(-1/2)+c,则x^2 f(x^3+1) dx为多少,求讲解

答：
∫f（x）dx=x^2/(1+x)^(-1/2)+C
∫x^2 f(x^3+1) dx
=(1/3)∫f(x^3+1)d(x^3+1)令t=x^3+1：
=(1/3)f(t)dt
=(1/3)*t^2/(1+t)^(-1/2)+C
=(1/3)(x^3+1)^2/(x^3+2)^(-1/2)+C