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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

Add a constant to the solution

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

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