What is the integral of 7/(x^2-9) ?

The integral of 7/(x^2-9) is -7/6 (ln| x/3+1|-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 7/(x^2-9)

\int \frac{7}{x ^{2} - 9} d x

- \frac{7}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

Solution steps

\int \frac{7}{x ^{2} - 9} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{x ^{2} - 9} d x

Apply Integral Substitution

= 7 \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

= 7 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 7 \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

= 7 \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}

= 7 \cdot \frac{1}{3} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{3}

= 7 \cdot \frac{1}{3} (- (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2}))

Simplify

7 \cdot \frac{1}{3} (- (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2})) : - \frac{7}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

= - \frac{7}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

Add a constant to the solution

= - \frac{7}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

\frac{\partial}{\partial y} (4 x^{2} - y^{2} + 2 y)

\int cos (6 x) cos (4 x) d x

x (x - 9)^{2}

\int_{2}^{4} 5 x^{2} d x

f (x) = 4 - x^{2}, at x = - 4

What is the integral of 7/(x^2-9) ?
The integral of 7/(x^2-9) is -7/6 (ln| x/3+1|-ln| x/3-1|)+C