What is the integral of 1/(yln(y)) ?

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

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integral of 1/(yln(y))

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

ln ∣ ln (y) ∣ + C

Solution steps

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

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 (y)

= ln ∣ ln (y) ∣

Add a constant to the solution

= ln ∣ ln (y) ∣ + C

\frac{\partial}{\partial v} (u + 2 v)

\int_{1}^{2} \frac{v ^{6} + 4 v ^{8}}{v ^{5}} d v

y^{^{''}} + 4 y^{^{'}} - 16 y = 0

\frac{\partial}{\partial y} (3 x^{2} (1 + ln (y)))

x \to 0 lim (\frac{2 + x - 2}{x})