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

f (x) = 8 x + ln (x^{2}), at x = 1

\int \frac{x ^{6} - x ^{3} + 36}{x ^{4} + 6 x ^{2}} d x

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

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

\frac{\partial}{\partial x} (2 x y^{3} - 1)

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