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

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

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

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

Solution steps

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

Apply Integral Substitution

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

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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} 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 \frac{( n ^{n} )}{n !}

\int \frac{1}{z} d z

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

(- 8.2) (- 10.5)

y = (2 x - 1)^{2} ln (2 x - 1)

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