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

The integral from 1 to infinity of x^2ln(x) is diverges

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integral from 1 to infinity of x^2ln(x)

\int_{1}^{\infty} x^{2} ln (x) d x

diverges

Solution steps

\int_{1}^{\infty} x^{2} ln (x) d x

Compute the indefinite integral:

\int x^{2} ln (x) d x = \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + C

Compute the boundaries:

\int_{1}^{\infty} x^{2} ln (x) d x =

diverges

= diverges

(12 y^{2} - x \cdot y) d x + (x^{2}) d y = 0

\int (- 3 x + 4) \cdot e^{- x} d x

\frac{d}{d x} (\frac{3 x ^{2}}{1 + x ^{3}})

\int \frac{x + 7}{5 - x} d x

y^{^{'}} = x^{2} y^{3}

What is the integral from 1 to infinity of x^2ln(x) ?
The integral from 1 to infinity of x^2ln(x) is diverges