What is the integral of 1/((y-1)(y+1)) ?

The integral of 1/((y-1)(y+1)) is -1/2 ln|y+1|+1/2 ln|y-1|+C

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integral of 1/((y-1)(y+1))

\int \frac{1}{( y - 1 ) ( y + 1 )} d y

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

Solution steps

\int \frac{1}{( y - 1 ) ( y + 1 )} d y

Expand

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

= \int \frac{1}{y ^{2} - 1} d y

Factor

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

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

Take the constant out:

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

= - \int \frac{1}{- y ^{2} + 1} d y

Use the common integral:

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{x \cdot ( ln ( x ) )} d x

x \to 1 lim (- x^{2} + 4 x - 8)

\frac{3 t}{t}

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{5 x ^{\frac{4}{5}}})

\frac{\partial}{\partial x} (6 x cos (x) sin (y))

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