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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

\int x^{1} ln (x) d x = \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} + C

Compute the boundaries:

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

diverges

= diverges

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