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

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

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

\int_{1}^{\infty} \frac{1}{x ^{2} - x} d x

diverges

Solution steps

\int_{1}^{\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_{1}^{\infty} \frac{1}{x ^{2} - x} d x =

diverges

= diverges

\frac{d}{d x} (4 x^{3} + 4)

x \frac{d y}{d x} = 3 y

\int \frac{d}{1} (\frac{1}{x}) d x

n = 1 \sum \infty \frac{8}{( n ) ( n + 3 )}

\int u^{- 3} (9 - u^{3} + u^{6}) d u

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