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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{3}{2} x

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

Combine same powers :

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

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

Add a constant to the solution

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

\frac{d}{d x} (3 \cdot e^{3 x})

\int x^{2} - 5 x + 9 d x

1 + t e^{t}

\frac{d}{d x} (sin^{2} (x) + y)

1 - 1 (t)

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