What is the integral of 1/(9v^2-16) ?

The integral of 1/(9v^2-16) is -1/24 (ln| 3/4 v+1|-ln| 3/4 v-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/(9v^2-16)

\int \frac{1}{9 v ^{2} - 16} d v

- \frac{1}{24} (ln \frac{3}{4} v + 1 - ln \frac{3}{4} v - 1) + C

Solution steps

\int \frac{1}{9 v ^{2} - 16} d v

Apply Integral Substitution

= \int \frac{1}{12 ( 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}{12} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{12} \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}{12} (- \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}{12} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{3}{4} v

= \frac{1}{12} (- (\frac{ln \frac{3}{4} v + 1}{2} - \frac{ln \frac{3}{4} v - 1}{2}))

Simplify

\frac{1}{12} (- (\frac{ln \frac{3}{4} v + 1}{2} - \frac{ln \frac{3}{4} v - 1}{2})) : - \frac{1}{24} (ln \frac{3}{4} v + 1 - ln \frac{3}{4} v - 1)

= - \frac{1}{24} (ln \frac{3}{4} v + 1 - ln \frac{3}{4} v - 1)

Add a constant to the solution

= - \frac{1}{24} (ln \frac{3}{4} v + 1 - ln \frac{3}{4} v - 1) + C

3 x^{2} - 2

\frac{\partial}{\partial s} (s t)

x \to 0 lim (2 x^{2} sin (\frac{1}{x}))

\int \frac{x ^{5}}{6} d x

x \to 0 lim (ln (1 - 9 x))

What is the integral of 1/(9v^2-16) ?
The integral of 1/(9v^2-16) is -1/24 (ln| 3/4 v+1|-ln| 3/4 v-1|)+C