What is the integral of (18)/(x^2-1) ?

The integral of (18)/(x^2-1) is -18(1/2 ln|x+1|-1/2 ln|x-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (18)/(x^2-1)

\int \frac{18}{x ^{2} - 1} d x

- 18 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

Solution steps

\int \frac{18}{x ^{2} - 1} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{x ^{2} - 1} d x

Factor

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

= 18 \cdot \int \frac{1}{- ( - x ^{2} + 1 )} d x

Take the constant out:

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

= 18 (- \int \frac{1}{- x ^{2} + 1} d x)

Use the common integral:

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

= 18 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

Simplify

18 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : - 18 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= - 18 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Add a constant to the solution

= - 18 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

n = 2 \sum \infty \frac{1}{ln ( n )}

csc^{2} (x)

f (x) = 3 x^{2} + 5 x - 4, at x = - 0.6667

y = u + \frac{2}{u}, u = (2 x + 1)^{3}, x = 0

y^{^{''}} + y = (t^{2} + t) \cdot sin (t)

What is the integral of (18)/(x^2-1) ?
The integral of (18)/(x^2-1) is -18(1/2 ln|x+1|-1/2 ln|x-1|)+C