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

The integral of 1/(u^2-9) is -1/6 (ln| u/3+1|-ln| u/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/(u^2-9)

\int \frac{1}{u ^{2} - 9} d u

- \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1) + C

Solution steps

\int \frac{1}{u ^{2} - 9} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} - 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{3} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

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

Substitute back

v = \frac{u}{3}

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

Simplify

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

= - \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{u}{3} + 1 - ln \frac{u}{3} - 1) + C

\int (t) d t

\int_{0}^{3} 4 e^{- 3 x} d x

\int \frac{y}{( y + 4 ) ( 2 y - 1 )} d y

\int e^{x + 2 y} d y

\frac{\partial}{\partial y} (y ln (y))

What is the integral of 1/(u^2-9) ?
The integral of 1/(u^2-9) is -1/6 (ln| u/3+1|-ln| u/3-1|)+C