f(x)=lnx
d/dx(lnx),Basedonthedefinitionofderviative
=lim(h->0)[f(x+h)-f(x)]/h
=lim(h->0)[ln(x+h)-lnx]/h
=lim(h->0)(1/h)ln[(x+h)/x]
=lim(h->0)(1/h)(x/h)(h/x)ln(1+h/x)
=lim(h->0)(1/x)ln[(1+h/x)^(x/h)]
=(1/x)lim(h->0)ln{(1+h/x)^[1/(h/x)]}
=(1/x)*lne,Definitionofe=lim(x->0)(1+x)^(1/x)
=1/x
∵d/dx(lnx)=1/x,integratebothsides,wehave
∫d/dx(lnx)dx=∫(1/x)dx
∫d(lnx)=∫(1/x)dx
=>∫(1/x)dx=lnx+C
Whenx>1,itisobviousthat∫(1/x)dx=lnx+C
Whenx0and-dy=dx
∫(1/x)dx=∫1/(-y)(-dy)=∫1/ydy=lny+C=ln(-x)+Cforx