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

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

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integral of 2ln(x^x)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int ln (x^{x}) d x

Apply Integration By Parts

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

= 2 (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}

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

Simplify

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

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

Add a constant to the solution

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

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