What is the integral from 7 to infinity of 9/(v^2-v) ?

The integral from 7 to infinity of 9/(v^2-v) is -9(ln(6)-ln(7))

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integral from 7 to infinity of 9/(v^2-v)

\int_{7}^{\infty} \frac{9}{v ^{2} - v} d v

- 9 (ln (6) - ln (7))

Decimal Notation

1.38735 \dots

Solution steps

\int_{7}^{\infty} \frac{9}{v ^{2} - v} d v

Compute the indefinite integral:

\int \frac{9}{v ^{2} - v} d v = 9 (- ln ∣ v ∣ + ln ∣ v - 1 ∣) + C

Compute the boundaries:

\int_{7}^{\infty} \frac{9}{v ^{2} - v} d v = 0 - 9 (- ln (7) + ln (6))

= 0 - 9 (- ln (7) + ln (6))

Simplify

= - 9 (ln (6) - ln (7))

\frac{\partial}{\partial x} (e^{\frac{x + y}{x - y}})

(10.2) (19.14)

x \to 2 + lim (\frac{∣ 2 - x ∣}{2 - x})

\int \frac{8}{9 + 8 x} d x

y = 2 ln (2 x)

What is the integral from 7 to infinity of 9/(v^2-v) ?
The integral from 7 to infinity of 9/(v^2-v) is -9(ln(6)-ln(7))