What is the integral of 8/(x^2-1) ?

The integral of 8/(x^2-1) is -8(1/2 ln|x+1|-1/2 ln|x-1|)+C

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integral of 8/(x^2-1)

\int \frac{8}{x ^{2} - 1} d x

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

Solution steps

\int \frac{8}{x ^{2} - 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2} - 1} d x

Factor

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

= 8 \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

= 8 (- \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}

= 8 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

Simplify

8 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : - 8 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= - 8 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Add a constant to the solution

= - 8 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

\int 7 cos^{3} (x) d x

\frac{\partial}{\partial x} (8 y^{4} x^{7} - 7 y^{3} x^{6})

\frac{d y}{d x}, y = x^{2} + 4

\frac{d}{d t} (e^{t} sin (2 t))

\frac{d}{d x} (\frac{2}{( x - 2 ) ^{2}})

What is the integral of 8/(x^2-1) ?
The integral of 8/(x^2-1) is -8(1/2 ln|x+1|-1/2 ln|x-1|)+C