What is the integral of-xln(x) ?

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

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integral of-xln(x)

\int - x ln (x) d x

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

Solution steps

\int - x ln (x) d x

Take the constant out:

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

= - \int x ln (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (tan (x))

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

y^{^{'}} = x y - sin (x) e^{- x^{2}} y^{3}

\frac{d}{d x} (\frac{ln ( 3 )}{x})

y^{^{''}} + 6 y^{^{'}} + 4 y = 0

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