How do you integrate $\frac{ln x}{2 x}$ $(lnx)/(2x)$ ?

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Answer 1 · 2018-05-21T09:09:13

Is an inmediate integral. See below

We know that if $f (x) = ln x$ $f(x)=lnx$ 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, $I=1/2[(lnx)(lnx)-int(1/x*lnx)dx], i.e.,$

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

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

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

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

How do you integrate (lnx)/(2x)lnx2x(lnx)/(2x)?