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

The integral of 1/(3x^2-2) is -1/(2sqrt(6))(ln|sqrt(3/2)x+1|-ln|sqrt(3/2)x-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(3x^2-2)

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

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

Solution steps

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

Factor

3 x^{2} - 2 : - (- 3 x^{2} + 2)

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

Take the constant out:

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

= - \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 = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - \frac{1}{6} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{3}{2} x

= - \frac{1}{6} \frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2}

Simplify

- \frac{1}{6} \frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2} : - \frac{1}{2 6} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

= - \frac{1}{2 6} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

Add a constant to the solution

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

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