What is the integral of 1/(3+2x^2) ?

The integral of 1/(3+2x^2) is 1/(sqrt(6))arctan(sqrt(2/3)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/(3+2x^2)

\int \frac{1}{3 + 2 x ^{2}} d x

\frac{1}{6} arctan (\frac{2}{3} x) + C

Solution steps

\int \frac{1}{3 + 2 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \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}{6} arctan (u)

Substitute back

u = \frac{2}{3} x

= \frac{1}{6} arctan (\frac{2}{3} x)

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{6} arctan (\frac{2}{3} x)

Add a constant to the solution

= \frac{1}{6} arctan (\frac{2}{3} x) + C

\int (2) d x

\int (tan (x) + cot (x)) d x

x \to 0 + lim ((\frac{1}{x})^{x^{2}})

arccsc (x)

x \to 2 lim (x^{3} - 2 x^{2} + x + 1)

What is the integral of 1/(3+2x^2) ?
The integral of 1/(3+2x^2) is 1/(sqrt(6))arctan(sqrt(2/3)x)+C