What is the integral from 0 to infinity of 1/(x(x^2+1)) ?

The integral from 0 to infinity of 1/(x(x^2+1)) is diverges

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integral from 0 to infinity of 1/(x(x^2+1))

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

diverges

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

diverges

= diverges

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

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

\int_{0}^{π} (x cos (x)) d x

What is the integral from 0 to infinity of 1/(x(x^2+1)) ?
The integral from 0 to infinity of 1/(x(x^2+1)) is diverges