What is the integral of-yln(y) ?

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

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integral of-yln(y)

\int - y ln (y) d y

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

Solution steps

\int - y ln (y) d y

Take the constant out:

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

= - \int y ln (y) d y

Apply Integration By Parts

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

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

\int \frac{y}{2} d y = \frac{y ^{2}}{4}

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

Simplify

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

Add a constant to the solution

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

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