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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

diverges

= diverges

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2} + 1})

\frac{d y}{d x} = - 0.05 y

\frac{\partial}{\partial y} (2 x y^{4} - 4)

\int x^{3} (x^{2} + 1)^{8} d x

\frac{d}{d z} (x e^{\frac{y}{z}})

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