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

The integral from 6 to infinity of 1/(x^2+x) is -(ln(6)-ln(7))

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integral from 6 to infinity of 1/(x^2+x)

\int_{6}^{\infty} \frac{1}{x ^{2} + x} d x

- (ln (6) - ln (7))

Decimal Notation

0.15415 \dots

Solution steps

\int_{6}^{\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_{6}^{\infty} \frac{1}{x ^{2} + x} d x = 0 - (ln (6) - ln (7))

= 0 - (ln (6) - ln (7))

Simplify

= - (ln (6) - ln (7))

\frac{d y}{d t} = 2 - y

\frac{d y}{d x} = x^{4} (32 - x^{5})^{5}

(- 1.6) (1.5)

ln (x), x = 1

\int \frac{4 x - 2}{x ^{4} + x ^{2}} d x

What is the integral from 6 to infinity of 1/(x^2+x) ?
The integral from 6 to infinity of 1/(x^2+x) is -(ln(6)-ln(7))