What is the integral of ln(x^x) ?

The integral of ln(x^x) is 1/2 x^2ln(x)-1/2 x^2+(x^2)/4+C

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integral of ln(x^x)

\int ln (x^{x}) d x

\frac{1}{2} x^{2} ln (x) - \frac{1}{2} x^{2} + \frac{x ^{2}}{4} + C

Solution steps

\int ln (x^{x}) d x

Apply Integration By Parts

: ln (x^{x}) x - \int (ln (x) + 1) x d x

= ln (x^{x}) x - \int (ln (x) + 1) x d x

\int (ln (x) + 1) x d x = \frac{1}{2} x^{2} (ln (x) + 1) - \frac{x ^{2}}{4}

= ln (x^{x}) x - (\frac{1}{2} x^{2} (ln (x) + 1) - \frac{x ^{2}}{4})

Simplify

ln (x^{x}) x - (\frac{1}{2} x^{2} (ln (x) + 1) - \frac{x ^{2}}{4}) : \frac{1}{2} x^{2} ln (x) - \frac{1}{2} x^{2} + \frac{x ^{2}}{4}

= \frac{1}{2} x^{2} ln (x) - \frac{1}{2} x^{2} + \frac{x ^{2}}{4}

Add a constant to the solution

= \frac{1}{2} x^{2} ln (x) - \frac{1}{2} x^{2} + \frac{x ^{2}}{4} + C

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What is the integral of ln(x^x) ?
The integral of ln(x^x) is 1/2 x^2ln(x)-1/2 x^2+(x^2)/4+C