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

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

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

2 arctan (x) + C

Solution steps

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

Use the common integral:

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

= 2 arctan (x)

Add a constant to the solution

= 2 arctan (x) + C

\frac{\partial}{\partial x} (- t + x)

\int_{- \frac{π}{3}}^{\frac{π}{3}} 1

x^{\frac{5}{8}}

x \to - 1 lim ((x + 2 x^{4})^{5})

t y^{^{'}} = 2 y - 3 t^{3} cos (4 t)

What is the integral of (2/(x^2+1)) ?
The integral of (2/(x^2+1)) is 2arctan(x)+C