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

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

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

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

- \frac{x}{ln ( x )} + Ei (ln (x)) + C

Solution steps

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

Apply u-substitution

= \int \frac{e ^{u}}{u ^{2}} d u

Apply Integration By Parts

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

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

= - \frac{e ^{u}}{u} - (- Ei (u))

Substitute back

u = ln (x)

= - \frac{e ^{l n (x)}}{ln ( x )} - (- Ei (ln (x)))

Simplify

- \frac{e ^{l n (x)}}{ln ( x )} - (- Ei (ln (x))) : - \frac{x}{ln ( x )} + Ei (ln (x))

= - \frac{x}{ln ( x )} + Ei (ln (x))

Add a constant to the solution

= - \frac{x}{ln ( x )} + Ei (ln (x)) + C

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