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

Answer 1

William Abdalla

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#

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Answer 2

Ezekiel Alexander

# 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#.

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Answer 3

Emma Aldridge

To integrate (\frac{\ln(x)}{2x}), you can use integration by parts with (u = \ln(x)) and (dv = \frac{1}{2x}dx). Then, differentiate (u) to get (du) and integrate (dv) to get (v). Finally, apply the integration by parts formula: (\int u , dv = uv - \int v , du).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.