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Integral ( ln(ln(x)) / x dx )

We use substitution to solve this. But first, I'm going to rearrange this to show the substitution in action.

Integral ( ln(ln(x)) (1/x) dx )

Let t = ln(x). That means

dt = (1/x) dx

Replace ln(x) with t, and replace (1/x) dx with dt (as per our substitution).

Integral ( ln(t) dt )

We solve for this particular integral using integration by parts.

Let u = ln(t). dv = dt.

du = (1/t) dt. v = t

The formula for integration by parts is

uv - Integral ( v du ), so we get

t ln(t) - Integral ( t(1/t) dt )

t ln(t) - Integral ( 1 dt )

t ln(t) - t + C

Back-substitute t = ln(x).

ln(x) ln(ln(x)) - ln(x) + C

Lnx In Integrali

∫ ln (ln (x)) dx/ x

Let ln(x) = t

1/x dx = dt

The integral becomes:

∫ ln(t) dt

Integrate this by parts

dv=dt

v=t

u=ln(t)

du= 1/t

∫ u dv = uv - ∫ vdu

t ln(t) - ∫ t (1/t) dt

= t ln(t) - t + C

Relace t with ln(x)

= ln(x) ln( ln(x) ) - ln(x) + C

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RE:

Integral of [ln(lnx)]/[x] dx?

can u plz show the steps clearly?

thanks much appreciate :)

Use the substitution y = ln(x), dy = (1/x) dx

Then it becomes INT ln(ln(x))/x dx = INT ln(y) dy = y*(ln(y) - 1) + c

= ln(x)*[ ln(ln(x)) - 1] + c

(ln(x) - 1) ln(ln(x))