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

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

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

\int \frac{1}{x 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

f (x) = cos (4 x + 1)

\int_{- 2}^{0} (4 - x) d x

\frac{\partial}{\partial x} (\frac{e ^{x}}{1 + e ^{x}})

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\frac{\partial}{\partial x} (sin (π (4 x - 4 y)))

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