What is the integral of 1/(64-x^2) ?

The integral of 1/(64-x^2) is 1/16 (ln| x/8+1|-ln| x/8-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(64-x^2)

\int \frac{1}{64 - x ^{2}} d x

\frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1) + C

Solution steps

\int \frac{1}{64 - x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{8 ( - 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}{8} \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}{8} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{8}

= \frac{1}{8} (\frac{ln \frac{x}{8} + 1}{2} - \frac{ln \frac{x}{8} - 1}{2})

Simplify

\frac{1}{8} (\frac{ln \frac{x}{8} + 1}{2} - \frac{ln \frac{x}{8} - 1}{2}) : \frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1)

= \frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1)

Add a constant to the solution

= \frac{1}{16} (ln \frac{x}{8} + 1 - ln \frac{x}{8} - 1) + C

\int (2 x^{2} - x) d x

\int 17 e^{u} d u

\int x (x^{2} + 9)^{3} d x

\int \frac{x}{x ^{3} + 3 x ^{2} + 7 x + 5} d x

\int \frac{x}{( x + 1 )} d x

What is the integral of 1/(64-x^2) ?
The integral of 1/(64-x^2) is 1/16 (ln| x/8+1|-ln| x/8-1|)+C