What is the integral of [pln(p)] ?

The integral of [pln(p)] is 1/2 p^2ln(p)-(p^2)/4+C

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integral of [pln(p)]

\int [p ln (p)] d p

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

Solution steps

\int p ln (p) d p

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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What is the integral of [pln(p)] ?
The integral of [pln(p)] is 1/2 p^2ln(p)-(p^2)/4+C