Question

∫Inx/x^2 dx 求它的不定积分!分子是Inx,分母是x的平方

Answer 1

∫Inx/x^2 dx
=-∫lnxd(1/x)
=-(lnx*1/x-∫1/x*dlnx)
=-(lnx*1/x-∫1/x^2dx)
=-lnx*1/x-1/x+C
=-(lnx+1)/x+C