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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

diverges

= diverges

\int_{- \frac{π}{4}}^{0} x sec^{2} (x) d x

\int_{1}^{4} (x^{2} - x + 1) d x

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

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

\int_{- 3}^{3} x^{3} + 4 x + 1 d x

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