integral of 2pln(p)

解答

\int 2 p ln (p) d p

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

求解步骤

\int 2 p ln (p) d p

提出常数:

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

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

使用分布积分法

= 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})

化简

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}

解答补常数

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

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