What is the integral of (1+t^2)^{-1} ?

The integral of (1+t^2)^{-1} is arctan(t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (1+t^2)^{-1}

integral

(1 + t^{2})^{- 1}

arctan (t) + C

Solution steps

\int (1 + t^{2})^{- 1} d t

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Use the common integral:

\int \frac{1}{1 + t ^{2}} d t = arctan (t)

= arctan (t)

Add a constant to the solution

= arctan (t) + C

What is the integral of (1+t^2)^{-1} ?
The integral of (1+t^2)^{-1} is arctan(t)+C