设f(X)为连续函数,则积分∫(上限n,下限1/n)[(1-1/x²)f(1+1/x²)]dx=

Question

设f(X)为连续函数,则积分∫(上限n,下限1/n)[(1-1/x²)f(1+1/x²)]dx=

Answer 1

令t=1+1/x² 则关于t的积分上限是1+1/n² 积分下限是1+n²
x=(t-1)^(-1/2) dx=-1/2(t-1)^(-3/2) dt
1-1/x²=2-t
原极限=∫(上限,下限)[-1/2(2-t)(t-1)^(-3/2)f(t)]dt