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

The integral of-2/(1+x^2) is -2arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-2/(1+x^2)

\int - \frac{2}{1 + x ^{2}} d x

- 2 arctan (x) + C

Solution steps

\int - \frac{2}{1 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= - 2 arctan (x)

Add a constant to the solution

= - 2 arctan (x) + C

\int cos^{2} (x) + sin^{2} (x) d x

\int \frac{2}{200 + x} d x

\int \frac{x ^{2} \cdot arctan ( x )}{1 + x ^{2}} d x

\int sin (x) (2 + cos (x))^{2} d x

\int (arctan (t))^{2} d t