What is the integral of 1/((x)ln(x)) ?

The integral of 1/((x)ln(x)) is ln|ln(x)|+C

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integral of 1/((x)ln(x))

\int \frac{1}{( x ) ln ( x )} d x

ln ∣ ln (x) ∣ + C

Solution steps

\int \frac{1}{x ln ( x )} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = ln (x)

= ln ∣ ln (x) ∣

Add a constant to the solution

= ln ∣ ln (x) ∣ + C

\int \frac{8}{1 + 9 r ^{2}} d r

\frac{d}{d x} (\frac{( x ^{2} )}{x ^{2} + 4})

f (θ) = sec^{2} (θ)

\frac{\partial}{\partial y} (4 x y + 4 y)

n = 0 \sum \infty - 6 (\frac{1}{4})^{2 n}

What is the integral of 1/((x)ln(x)) ?
The integral of 1/((x)ln(x)) is ln|ln(x)|+C