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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{2}^{\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_{2}^{\infty} \frac{1}{x ^{2} - x} d x = 0 - (- ln (2))

= 0 - (- ln (2))

Simplify

= ln (2)

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