What is the integral of 2pln(p) ?

The integral of 2pln(p) is p^2ln(p)-(p^2)/2+C

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integral of 2pln(p)

\int 2 p ln (p) d p

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

Solution steps

\int 2 p ln (p) d p

Take the constant out:

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

= 2 \cdot \int p ln (p) d p

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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