题目

求定积分,上n下1/n ∫(1-1/x^2) f(1+1/x^2)dx=?,

提问时间：2021-02-05

答案

∫[1/n,n](1-1/x^2)f(1+1/x^2)dx
=∫[1/n,n]f(1+1/x^2)d(x+1/x)
x+1/x=u f(1+1/x^2)=g(u)
x=n,u=n+1/n x=1/n u=n+1/n
=∫[n+1/n,n+1/n]g(u)du
=0