What is the integral of (12)/((x^2-9)) ?

The integral of (12)/((x^2-9)) is -4(1/2 ln| x/3+1|-1/2 ln| x/3-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (12)/((x^2-9))

\int \frac{12}{( x ^{2} - 9 )} d x

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

Solution steps

\int \frac{12}{x ^{2} - 9} d x

Take the constant out:

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

= 12 \cdot \int \frac{1}{x ^{2} - 9} d x

Apply Integral Substitution

= 12 \cdot \int \frac{1}{3 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 12 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 12 \cdot \frac{1}{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

= 12 \cdot \frac{1}{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}

= 12 \cdot \frac{1}{3} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{3}

= 12 \cdot \frac{1}{3} (- (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2}))

Simplify

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

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

Add a constant to the solution

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

\int \frac{w ^{5}}{8 w ^{2} + 1} d w

f (x) = 3 x - 2 x^{2}, (- 2, - 14)

\frac{\partial}{\partial x} (- 6 x y - x^{6} - y^{6})

\int \frac{1}{6 x + 5} d x

\int 3 x + 3 d x

What is the integral of (12)/((x^2-9)) ?
The integral of (12)/((x^2-9)) is -4(1/2 ln| x/3+1|-1/2 ln| x/3-1|)+C