What is the integral from 3 to infinity of 6/(x^2-x) ?

The integral from 3 to infinity of 6/(x^2-x) is -6(ln(2)-ln(3))

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integral from 3 to infinity of 6/(x^2-x)

\int_{3}^{\infty} \frac{6}{x ^{2} - x} d x

- 6 (ln (2) - ln (3))

Decimal Notation

2.43279 \dots

Solution steps

\int_{3}^{\infty} \frac{6}{x ^{2} - x} d x

Compute the indefinite integral:

\int \frac{6}{x ^{2} - x} d x = 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + C

Compute the boundaries:

\int_{3}^{\infty} \frac{6}{x ^{2} - x} d x = 0 - 6 (- ln (3) + ln (2))

= 0 - 6 (- ln (3) + ln (2))

Simplify

= - 6 (ln (2) - ln (3))

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\int \frac{3}{1 + x ^{2}} d x

What is the integral from 3 to infinity of 6/(x^2-x) ?
The integral from 3 to infinity of 6/(x^2-x) is -6(ln(2)-ln(3))