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

The integral from 1 to infinity of t/(1+t^2) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{1}^{\infty} \frac{t}{1 + t ^{2}} d t

diverges

Solution steps

\int_{1}^{\infty} \frac{t}{1 + t ^{2}} d t

Compute the indefinite integral:

\int \frac{t}{1 + t ^{2}} d t = \frac{1}{2} ln 1 + t^{2} + C

Compute the boundaries:

\int_{1}^{\infty} \frac{t}{1 + t ^{2}} d t =

diverges

= diverges

\int_{0}^{11} t^{2} e^{2 t} d t

\int_{0}^{1} e^{x^{2}} 2 x d x

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

\int_{- 2}^{4} x^{2} sin (x^{3} + 3) d x

\int_{0}^{3} 5 e^{x} d x

What is the integral from 1 to infinity of t/(1+t^2) ?
The integral from 1 to infinity of t/(1+t^2) is diverges