What is the integral from-7 to infinity of 1/(x+9) ?

The integral from-7 to infinity of 1/(x+9) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-7 to infinity of 1/(x+9)

\int_{- 7}^{\infty} \frac{1}{x + 9} d x

diverges

Solution steps

\int_{- 7}^{\infty} \frac{1}{x + 9} d x

Compute the indefinite integral:

\int \frac{1}{x + 9} d x = ln ∣ x + 9 ∣ + C

Compute the boundaries:

\int_{- 7}^{\infty} \frac{1}{x + 9} d x =

diverges

= diverges

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

\int_{2}^{3} \frac{1}{x ( x - 1 )} d x

\int_{0}^{3} (2 - x)^{2} d x

\int_{x^{2}}^{x^{3}} sin (x) d x

\int_{0}^{l n (13)} tanh (x) d x

What is the integral from-7 to infinity of 1/(x+9) ?
The integral from-7 to infinity of 1/(x+9) is diverges