What is the integral from 3 to infinity of 1/(x^2+x) ?

The integral from 3 to infinity of 1/(x^2+x) is -(ln(3)-2ln(2))

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integral from 3 to infinity of 1/(x^2+x)

\int_{3}^{\infty} \frac{1}{x ^{2} + x} d x

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

Decimal

0.28768 \dots

Solution steps

\int_{3}^{\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_{3}^{\infty} \frac{1}{x ^{2} + x} d x = 0 - (ln (3) - 2 ln (2))

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

Simplify

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

What is the integral from 3 to infinity of 1/(x^2+x) ?
The integral from 3 to infinity of 1/(x^2+x) is -(ln(3)-2ln(2))