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

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

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integral of 1/x ln(1/x)

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

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

Solution steps

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

Simplify

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

= \int - \frac{ln ( x )}{x} d x

Take the constant out:

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

= - \int \frac{ln ( x )}{x} d x

Apply u-substitution

: \int u d u

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = ln (x)

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

Add a constant to the solution

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

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