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

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

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

- \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣ + C

Solution steps

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

Factor

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

= \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

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣ + C

x \cdot 5 x^{2}

\int \frac{1}{1 + cos ^{2} ( θ )} d θ

\frac{d}{d x} (arccos (sin (x^{2} + 9 x)))

f (x) = 3 x + 4

x \to 0 lim ((1 + \frac{3}{x})^{x})

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