What is the integral from 5 to infinity of 1/(x^2+x) ?

The integral from 5 to infinity of 1/(x^2+x) is -(ln(5)-ln(6))

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integral from 5 to infinity of 1/(x^2+x)

\int_{5}^{\infty} \frac{1}{x ^{2} + x} d x

- (ln (5) - ln (6))

Decimal Notation

0.18232 \dots

Solution steps

\int_{5}^{\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_{5}^{\infty} \frac{1}{x ^{2} + x} d x = 0 - (ln (5) - ln (6))

= 0 - (ln (5) - ln (6))

Simplify

= - (ln (5) - ln (6))

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What is the integral from 5 to infinity of 1/(x^2+x) ?
The integral from 5 to infinity of 1/(x^2+x) is -(ln(5)-ln(6))