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

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

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

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

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

Solution steps

\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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x \to 3 lim ((x - 3)^{2})

\frac{d}{d x} (e^{- 3 x^{4}} \cdot ln (7 x^{2}))

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