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

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

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integral of 1/((y^2-1))

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

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

Solution steps

\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

cos^{3} (x) + sin^{4} (x)

\int \frac{3 x - 2}{x + 1} d x

\frac{\partial}{\partial x} (\frac{2 x ^{2}}{( 2 x + 7 y ) ^{2}})

x \to - 3 lim (∣ - x - 1 ∣)

\int \frac{1}{( x - 3 ) ( x - 1 )} d x

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