What is the integral from 1 to 2 of p^2ln(p) ?

The integral from 1 to 2 of p^2ln(p) is -7/9+8/3 ln(2)

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integral from 1 to 2 of p^2ln(p)

\int_{1}^{2} p^{2} ln (p) d p

- \frac{7}{9} + \frac{8}{3} ln (2)

Decimal Notation

1.07061 \dots

Solution steps

\int_{1}^{2} p^{2} ln (p) d p

Apply Integration By Parts

: [\frac{1}{3} p^{3} ln (p) - \int \frac{p ^{2}}{3} d p]_{1}^{2}

= [\frac{1}{3} p^{3} ln (p) - \int \frac{p ^{2}}{3} d p]_{1}^{2}

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

= [\frac{1}{3} p^{3} ln (p) - \frac{p ^{3}}{9}]_{1}^{2}

Compute the boundaries:

- \frac{7}{9} + \frac{8}{3} ln (2)

= - \frac{7}{9} + \frac{8}{3} ln (2)

\int_{2}^{3} x^{3} - 4 x d x

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\int_{- 1}^{2} x - 1 d x

\int_{0}^{6} \frac{1}{( x - 9 ) ^{\frac{2}{3}}} d x

What is the integral from 1 to 2 of p^2ln(p) ?
The integral from 1 to 2 of p^2ln(p) is -7/9+8/3 ln(2)