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

f (x) = 5 x^{3}, at x = 2

\frac{\partial}{\partial x} (x y + cos (x))

\frac{d}{d x} f (x) = (2 x - 6)^{2} (x^{2} + x + 1)^{2}

\frac{d}{d x} (\frac{4}{sin ( x ) + cos ( x )})

\int cos (x) - 9 sin (x) d 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