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

(e^{4 x})^{^{'}}

\frac{d}{d x} (5 \cdot e^{x} + 2 \cdot 3^{x})

e^{- t^{2}}

\int (x^{2} e^{x}) d x

y = x^{3} + 1, (0, 2)

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