How do you integrate #(lnx)/(2x)#?

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Answer 1 · 2018-05-21T09:09:13

Is an inmediate integral. See below

We know that if #f(x)=lnx# then #f´(x)=1/x# and with the chain rule

if #f(x)=ln(g(x))# then #f´(x)=(g´(x))/g(x)#

Then #intlnx/(2x)dx=1/4ln^2x+C#

Answer 2 · 2018-05-21T09:15:55

# 1/4(lnx)^2+C#.

Let, #I=intlnx/(2x)dx=1/2intlnx*1/xdx#.

Now, we use IBP : #intu*v'dx=u*v-intu'*vdx#, with,

#u=lnx and v'=1/x#.

#rArr u'=1/x and v=int1/xdx=lnx#.

Hence, #I=1/2[(lnx)(lnx)-int(1/x*lnx)dx], i.e., #

#I=1/2(lnx)^2-1/2intlnx*1/xdx, or, #

# I=1/2(lnx)^2-I#.

# :. 2I=1/2(lnx)^2#.

# rArr I=1/4(lnx)^2+C#.