What is the integral of 6/(1-x^2) ?

The integral of 6/(1-x^2) is 6((ln|x+1|)/2-(ln|x-1|)/2)+C

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integral of 6/(1-x^2)

\int \frac{6}{1 - x ^{2}} d x

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

Solution steps

\int \frac{6}{1 - x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{1 - x ^{2}} d x

Use the common integral:

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

= 6 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})

Add a constant to the solution

= 6 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}) + C

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What is the integral of 6/(1-x^2) ?
The integral of 6/(1-x^2) is 6((ln|x+1|)/2-(ln|x-1|)/2)+C