What is the integral of 1/(2x^2-18) ?

The integral of 1/(2x^2-18) is -1/12 (ln| x/3+1|-ln| x/3-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2x^2-18)

\int \frac{1}{2 x ^{2} - 18} d x

- \frac{1}{12} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

Solution steps

\int \frac{1}{2 x ^{2} - 18} 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

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{6} \cdot \int \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}{6} (- \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{2}{3 2} x

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

Simplify

\frac{1}{6} - \frac{ln \frac{2}{3 2} x + 1}{2} - \frac{ln \frac{2}{3 2} x - 1}{2} : - \frac{1}{12} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

= - \frac{1}{12} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

Add a constant to the solution

= - \frac{1}{12} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

\int (0.3 t^{2} + 0.02 t + 2) d t

\int 2 x^{3} (x^{2} + 1)^{65} d x

\int \frac{1}{17 x} d x

\int tan^{3} (4 x) sec^{3} (4 x) d x

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

What is the integral of 1/(2x^2-18) ?
The integral of 1/(2x^2-18) is -1/12 (ln| x/3+1|-ln| x/3-1|)+C