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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

= 0 - (- ln (2))

Simplify

= ln (2)

\frac{d}{d x} (2^{x} ln^{2} (2))

9 x + 11, 7 x + 14, [23, 13]

\frac{\partial}{\partial y} (e^{y} sin (x))

\frac{( 4 s - 18 )}{s ^{2} - 6 s + 10}

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