What is the integral of (ln(ln(k)))/(2k) ?

The integral of (ln(ln(k)))/(2k) is 1/2 (ln(k)ln(ln(k))-ln(k))+C

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integral of (ln(ln(k)))/(2k)

\int \frac{ln ( ln ( k ) )}{2 k} d k

\frac{1}{2} (ln (k) ln (ln (k)) - ln (k)) + C

Solution steps

\int \frac{ln ( ln ( k ) )}{2 k} d k

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{ln ( ln ( k ) )}{k} d k

Apply u-substitution

: \int ln (u) d u

= \frac{1}{2} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} (u ln (u) - u)

Substitute back

u = ln (k)

= \frac{1}{2} (ln (k) ln (ln (k)) - ln (k))

Add a constant to the solution

= \frac{1}{2} (ln (k) ln (ln (k)) - ln (k)) + C

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What is the integral of (ln(ln(k)))/(2k) ?
The integral of (ln(ln(k)))/(2k) is 1/2 (ln(k)ln(ln(k))-ln(k))+C