Integration

it's logx + c

i see no other method to do that. by parts give I = I,
so i can't thnk of any othe rmethod

I found only one method.
Let y=e^x ;(x=log_ey)
dydx=e^x
dydx=y
dyy=dx
∫dyy=x+c
∫dyy=log_ey+c

y→x
∫dxx=log_ex+c

where c is a constant

Any more ideas ?

one more

y=lnx

dy/dx = 1/x

integrating we have

y+c=∫1/x
so
∫1/x =lnx + c

Using y=lnx

dy/dx = 1/x

implies we already know ∫dxx=lnx +c

yes,thats correct.solve it

plz tell how to post a latexified equation