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

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

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