设不定积分1～x^2 f（t）dt=lnx，则f（x）=

Question

如题所述

Answer 1

f(x)=∫0→x dt/(1+t^2) +∫0→1/x dt/(1+t^2)
=arctant|(0→x )+arctant|(0→1/x)
=arctanx+arctan(1/x)
=arctanx+arccotx
=π/2
或者f'(x)=1/(1+x^2)+1/(1+1/x^2)*(-1/x^2)
=1/(1+x^2)-1/(1+x^2)=0
故f(x)=c=f(1)=2∫0→1 dt/(1+t^2)
=2*(arctan1-arctan0)
=π/2本回答被网友采纳