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

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

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

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

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Take the constant out:

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

= - 2 \cdot \int \frac{1}{x ^{2} - 1} d x

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

x \to 0 + lim (\frac{x ^{2} - 15}{x})

\frac{d}{d x} (x^{2} sin (6 x))

y^{^{''}} + 9 y = x e^{2 x}

\frac{\partial}{\partial x} (\frac{3}{( x ^{2} + z ^{2} )})

\frac{d}{d z} (x e^{z})

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