What is the integral of 6/(1+x^2) ?

The integral of 6/(1+x^2) is 6arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6/(1+x^2)

\int \frac{6}{1 + x ^{2}} d x

6 arctan (x) + C

Solution steps

\int \frac{6}{1 + x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 6 arctan (x)

Add a constant to the solution

= 6 arctan (x) + C

x \to 1 lim (f (x) + g (x))

n = 2 \sum \infty \frac{n}{2 n ^{2} - 3}

\frac{\partial}{\partial x} (e^{- x y} + y^{3} x^{4})

\frac{d}{d x} (3 x^{3} + 2 x^{2} - 1 x + 5)

\int_{0}^{3} e^{- x^{2}} d x