What is the integral of wln(w) ?

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

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integral of wln(w)

\int w ln (w) d w

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

Solution steps

\int w ln (w) d w

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

y^{^{'}} + \frac{3}{2 x} y = 0

\int_{1}^{5} (4 x - 1) d x

\int ln (x y) d x

\int_{- \infty}^{10} x e^{\frac{x}{5}} d x

x y^{^{'}} - y + 1 = 0

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