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

The integral of 1/(ln(y)) is \Ei(ln(y))+C

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integral of 1/(ln(y))

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

Ei (ln (y)) + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{u} d u

= \int \frac{e ^{u}}{u} d u

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= Ei (u)

Substitute back

u = ln (y)

= Ei (ln (y))

Add a constant to the solution

= Ei (ln (y)) + C

\frac{d}{d x} (8 x + 1)

\int_{0}^{2} (x + 2)^{2} d x

f (x) = 6 x

\frac{d}{d x} (sinh (ln (x)))

n \to \infty lim (a n^{2})