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))

integral

\frac{1}{ln ( 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

- t^{2} + 2

(e^{- x})^{2} (x^{3})

x^{2} + y^{2}

\frac{t}{( 1 + t )}

\frac{ln ( 1 )}{x}