What is the integral from 8 to infinity of 1/(x^2+x) ?

The integral from 8 to infinity of 1/(x^2+x) is -(3ln(2)-2ln(3))

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integral from 8 to infinity of 1/(x^2+x)

\int_{8}^{\infty} \frac{1}{x ^{2} + x} d x

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

Decimal

0.11778 \dots

Solution steps

\int_{8}^{\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_{8}^{\infty} \frac{1}{x ^{2} + x} d x = 0 - (3 ln (2) - 2 ln (3))

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

Simplify

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

y^{^{'}} = (y + 4 x)^{2}

x \to 1 + lim (\frac{x ^{2}}{x ^{2} - 1})

\frac{d y}{d x} = 3 e^{2 x} y^{2}

ln (x + 2)

\int x^{3} 5^{- x} d x

What is the integral from 8 to infinity of 1/(x^2+x) ?
The integral from 8 to infinity of 1/(x^2+x) is -(3ln(2)-2ln(3))