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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 3 to infinity of 1/(x^2+x)

\int_{3}^{\infty} \frac{1}{x ^{2} + x} d x

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

Decimal Notation

0.28768 \dots

Solution steps

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

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

Simplify

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

\frac{d}{d x} ((3 x - 2)^{4})

x^{7} sin (x)

\int 2 x (x^{2} + 6)^{9} d x

\frac{d}{d x} (arcsin (4 - 3 x))

\int_{0}^{\frac{1}{2}} \frac{3}{1 + 4 x ^{2}} d x

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