What is the integral from-infinity to 0 of 1/(3-6x) ?

The integral from-infinity to 0 of 1/(3-6x) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-infinity to 0 of 1/(3-6x)

\int_{- \infty}^{0} \frac{1}{3 - 6 x} d x

diverges

Solution steps

\int_{- \infty}^{0} \frac{1}{3 - 6 x} d x

Compute the indefinite integral:

\int \frac{1}{3 - 6 x} d x = - \frac{1}{6} ln ∣ 3 - 6 x ∣ + C

Compute the boundaries:

\int_{- \infty}^{0} \frac{1}{3 - 6 x} d x =

diverges

= diverges

y = e^{x}, y = e^{2 x}, x = ln (4)

(cos (\frac{1}{x}))^{^{'}}

\int \frac{30 x ^{2} - 49}{x ^{4}} d x

\frac{\partial}{\partial x} (\frac{x}{x - 1})

x \to 0 lim (2 x^{2} cos (\frac{5}{x}))

What is the integral from-infinity to 0 of 1/(3-6x) ?
The integral from-infinity to 0 of 1/(3-6x) is diverges