What is the integral of 1/(2+4x^2) ?

The integral of 1/(2+4x^2) is 1/(2sqrt(2))arctan(sqrt(2)x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2+4x^2)

\int \frac{1}{2 + 4 x ^{2}} d x

\frac{1}{2 2} arctan (2 x) + C

Solution steps

\int \frac{1}{2 + 4 x ^{2}} d x

Apply Integral Substitution

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

Take the constant out:

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

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

\frac{1}{2 + 2 u ^{2}} = \frac{1}{2} \frac{1}{1 + u ^{2}}

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} arctan (u)

Substitute back

u = \frac{2}{2} x

= \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{2}{2} x)

Simplify

\frac{1}{2} \cdot \frac{1}{2} arctan (\frac{2}{2} x) : \frac{1}{2 2} arctan (2 x)

= \frac{1}{2 2} arctan (2 x)

Add a constant to the solution

= \frac{1}{2 2} arctan (2 x) + C

n = 0 \sum \infty sin (π \cdot n)

\frac{d}{d x} (6 x x)

x \to - \infty lim (3^{x} x)

\frac{d}{d t} (2 sin (t))

\frac{d x}{d t} = t^{2} x

What is the integral of 1/(2+4x^2) ?
The integral of 1/(2+4x^2) is 1/(2sqrt(2))arctan(sqrt(2)x)+C