What is the integral of 1/(k(ln(k))^2) ?

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

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integral of 1/(k(ln(k))^2)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = ln (k)

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

Add a constant to the solution

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

x \to 0 lim (\frac{1}{sin ( \frac{1}{x} )})

\frac{d}{d x} ((2 x - 1)^{4})

\frac{d}{d q} (- q^{2} + 36)

\frac{s + 2}{( s + 3 ) ^{2} ( s - 1 )}

x \to 0 - lim (\frac{3}{1 + e ^{\frac{1}{x}}})

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