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

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

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

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

Solution steps

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

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{v}{8}

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

Simplify

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

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

Add a constant to the solution

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

\int 7 x^{3} e^{x^{2}} d x

\int \frac{x ^{5} + 1}{x ^{3} ( x + 2 )} d x

\int \frac{( ln ( x ) ) ^{2}}{( x )} d x

\int \frac{3 ^{2 x}}{4 + 3 ^{2 x}} d x

\int \frac{8 x ^{5} + 4 x}{2 x ^{2}} d x

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