What is the integral of 4/(x^2-1) ?

The integral of 4/(x^2-1) is -4(1/2 ln|x+1|-1/2 ln|x-1|)+C

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integral of 4/(x^2-1)

\int \frac{4}{x ^{2} - 1} d x

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

Solution steps

\int \frac{4}{x ^{2} - 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} - 1} d x

Factor

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

= 4 \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

= 4 (- \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}

= 4 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

Simplify

4 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : - 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= - 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Add a constant to the solution

= - 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

\frac{d}{d x} (- \frac{1}{2} sin (x))

\int \frac{x ^{2}}{a ^{2} - x ^{2}} d x

e^{x}, e^{- x}, [- 1, 1]

\int \frac{1}{( x ^{2} - 121 ) ^{\frac{3}{2}}} d x

8 x^{6} + 5 w^{2} + 7 w

What is the integral of 4/(x^2-1) ?
The integral of 4/(x^2-1) is -4(1/2 ln|x+1|-1/2 ln|x-1|)+C