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

The integral of (x^{-1})/((ln(x))^2) is -1/(ln(x))+C

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integral of (x^{-1})/((ln(x))^2)

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

- \frac{1}{ln ( x )} + C

Solution steps

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

Simplify

\frac{x ^{- 1}}{( ln ( x ) ) ^{2}} : \frac{1}{x ln ^{2} ( x )}

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = ln (x)

= - \frac{1}{ln ( x )}

Add a constant to the solution

= - \frac{1}{ln ( x )} + C

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