What is the integral of-x*ln(x) ?

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

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integral of-x*ln(x)

\int - x \cdot 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

\int \frac{1}{1 + 16 x ^{2}} d x

\int_{- 1}^{2} (- y^{2} + y + 2) d y

x^{3} cos (x^{2})

\int \frac{x ^{2}}{25 - 9 x ^{2}} d x

2 y^{^{''}} - 9 y^{^{'}} - 5 y = 0

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