What is the integral from 1 to infinity of xln(x) ?

The integral from 1 to infinity of xln(x) is diverges

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integral from 1 to infinity of xln(x)

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

diverges

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

diverges

= diverges

\int_{- 1}^{1} \frac{1}{1 - ∣ x ∣} d x

\int_{1}^{5} x e^{x} d x

\int_{0}^{1} \frac{1}{x ( x ^{2} + 1 )} d x

\int_{0}^{2 π} sin^{2} (t) cos (t) d t

\int_{0}^{1} (x^{2} + 2 x + 1) d x

What is the integral from 1 to infinity of xln(x) ?
The integral from 1 to infinity of xln(x) is diverges