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

f (x) = - \frac{3}{x ^{2}}

\int 1 + (sinh (x))^{2} d x

\int 2 e^{- 2 y} d y

\frac{d}{d x} (e^{g (x) (x^{3}) - 2})

y^{^{''}} + 4 y^{^{'}} + 6 y = 0, y (0) = - 4, y^{^{'}} (0) = - 1

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