Help, how do I prove this integration formula, check my work

I'm not sure if I understand this question(s) correctly. How do I prove the integration formula,

∫ln|u|du = u(-1+ln|u|)+c,
by integrating ∫ln|x|dx = x(-1+ln|x|)+c

Do I use integration by parts?
Using integration by parts, I have u=ln|x|, dv=dx, du=dx/x and v=x
Substituting in to Uv-∫vdu, I have,
=x(ln|x|-1)+c

Then using my result, prove the integration formula,

∫(lnu)^2 du = u[ 2 - 2lnu + (lnu)^2] + C
∫ln^2xdx = x (2 - 2lnx + (ln^2) * x) + C

Using integration by parts again I have, u=ln^2(x), dv=dx, du= ( ( 2ln|x| ) / x ) dx, v=x
and substituting and using integration by parts again on the last part I have,
= x(ln^2x - 2ln(x) + 2) + C

I'm not sure what is meant from using the first formula to prove the second.

"I'm not sure what is meant from using the first formula to prove the second."

When you do integration by parts the second time in part 2, you're just repeating the integration you did in part 1. So just use result from part 1 instead, and avoid the aggravation of integrating by parts yet again.

1. You've shown that ∫ ln|x| dx = x (ln|x|−1) + c

2. Show ∫ (lnx)² dx = x (2 − 2lnx + (lnx)²) + C by using result from 1.

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u = (lnx)² . . . . . . . dv = dx
du = 2 lnx/x dx . . . . v = x

∫ u dv = u v − ∫ v du
∫ (lnx)² dx = x (lnx)² − ∫ 2x lnx/x dx
. . . . . . . . = x ln²x − 2 ∫ lnx dx ----> use result from 1 to find integral
. . . . . . . . = x ln²x − 2 x (lnx − 1) + C
. . . . . . . . = x (ln²x − 2 lnx + 2) + C

indeed you must use integration by parts.

∫ lnx dx

put lnx= u and dx=dv

dx=dv gives us that x=v

and u ' = 1/x

the integration by parts formula is:

u*v - ∫ u ' * v

put our details:

lnx *x - ∫ (1/x)*x dx

xlnx - ∫ 1 dx

xlnx -x +c => x(-1+lnx) +c

Yes , you are correct , u=ln^2(x), dv=dx ...