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

The integral of w^2ln(w) is 1/3 w^3ln(w)-(w^3)/9+C

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integral of w^2ln(w)

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

\frac{1}{3} w^{3} ln (w) - \frac{w ^{3}}{9} + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{3} w^{3} ln (w) - \int \frac{w ^{2}}{3} d w

= \frac{1}{3} w^{3} ln (w) - \int \frac{w ^{2}}{3} d w

\int \frac{w ^{2}}{3} d w = \frac{w ^{3}}{9}

= \frac{1}{3} w^{3} ln (w) - \frac{w ^{3}}{9}

Add a constant to the solution

= \frac{1}{3} w^{3} ln (w) - \frac{w ^{3}}{9} + C

2 x + 4

\frac{1}{2 t ^{2} + 1}

x \to 0 lim (x^{2} + 1)

\int a^{x} e^{x} d x

y^{^{'}} - 5 y = - t

What is the integral of w^2ln(w) ?
The integral of w^2ln(w) is 1/3 w^3ln(w)-(w^3)/9+C