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

The integral of 1/(x^2ln(x)) is \Ei(-ln(x))+C

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

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

Ei (- ln (x)) + C

Solution steps

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

Apply u-substitution

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{u} u} = e^{- u} \frac{1}{u}

= \int e^{- u} \frac{1}{u} d u

Simplify

e^{- u} \frac{1}{u} : \frac{e ^{- u}}{u}

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

Apply u-substitution

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

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

Use the nonelementary integral:

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

= Ei (v)

Substitute back

= Ei (- ln (x))

Add a constant to the solution

= Ei (- ln (x)) + C

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