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

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

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

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

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

Decimal Notation

0.40546 \dots

Solution steps

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

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

Simplify

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

\frac{d}{d x} (x^{2} lo g_{e} (x))

\int \frac{x - 3}{x ^{2} + 49} d x

x \to 0 lim (\frac{x}{arctan ( x )})

\frac{4}{1 + x ^{2}}

f (x) = cos (x) - x sin (x)

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