What is the integral from 3 to infinity of 8/(x^2-x) ?

The integral from 3 to infinity of 8/(x^2-x) is -8(ln(2)-ln(3))

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integral from 3 to infinity of 8/(x^2-x)

\int_{3}^{\infty} \frac{8}{x ^{2} - x} d x

- 8 (ln (2) - ln (3))

Solution steps

\int_{3}^{\infty} \frac{8}{x ^{2} - x} d x

Compute the indefinite integral:

\int \frac{8}{x ^{2} - x} d x = 8 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + C

Compute the boundaries:

\int_{3}^{\infty} \frac{8}{x ^{2} - x} d x = 0 - 8 (- ln (3) + ln (2))

= 0 - 8 (- ln (3) + ln (2))

Simplify

= - 8 (ln (2) - ln (3))

F (s) = \frac{6}{s ^{2} + 4}

\int - 6 sin (2 x) + 3 cos (3 x) d x

\frac{d y}{d x} = \frac{15 x}{y}

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

What is the integral from 3 to infinity of 8/(x^2-x) ?
The integral from 3 to infinity of 8/(x^2-x) is -8(ln(2)-ln(3))