What is the integral of 1/(x^2-48) ?

The integral of 1/(x^2-48) is -1/(8sqrt(3))(ln| x/(4sqrt(3))+1|-ln| x/(4sqrt(3))-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x^2-48)

\int \frac{1}{x ^{2} - 48} d x

- \frac{1}{8 3} (ln \frac{x}{4 3} + 1 - ln \frac{x}{4 3} - 1) + C

Solution steps

\int \frac{1}{x ^{2} - 48} d x

Apply Integral Substitution

= \int \frac{1}{4 3 ( 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}{4 3} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{4 3} \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}{4 3} (- \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}{4 3} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{4 3}

= \frac{1}{4 3} - \frac{ln \frac{x}{4 3} + 1}{2} - \frac{ln \frac{x}{4 3} - 1}{2}

Simplify

\frac{1}{4 3} - \frac{ln \frac{x}{4 3} + 1}{2} - \frac{ln \frac{x}{4 3} - 1}{2} : - \frac{1}{8 3} (ln \frac{x}{4 3} + 1 - ln \frac{x}{4 3} - 1)

= - \frac{1}{8 3} (ln \frac{x}{4 3} + 1 - ln \frac{x}{4 3} - 1)

Add a constant to the solution

= - \frac{1}{8 3} (ln \frac{x}{4 3} + 1 - ln \frac{x}{4 3} - 1) + C

What is the integral of 1/(x^2-48) ?
The integral of 1/(x^2-48) is -1/(8sqrt(3))(ln| x/(4sqrt(3))+1|-ln| x/(4sqrt(3))-1|)+C