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

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

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

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

- \frac{1}{ln ( 2 x )} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

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

Add a constant to the solution

= - \frac{1}{ln ( 2 x )} + C

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