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

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

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integral of 6/(x^2-1)

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

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

Solution steps

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

Factor

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

= 6 \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

= 6 (- \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}

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 4 x^{2} sin (x) - 7 x

\int_{0}^{5} \frac{1}{3 x + 1} d x

x \to - 1 - lim (\frac{∣ x + 1 ∣}{x + 1})

\int 9 (2 x - 9)^{8} d x

(1 + x)^{n}

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