What is the integral of 5/(x^2-1) ?

The integral of 5/(x^2-1) is -5(1/2 ln|x+1|-1/2 ln|x-1|)+C

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integral of 5/(x^2-1)

\int \frac{5}{x ^{2} - 1} d x

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

Solution steps

\int \frac{5}{x ^{2} - 1} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} - 1} d x

Factor

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

= 5 \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

= 5 (- \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}

= 5 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

Simplify

5 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Add a constant to the solution

= - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

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

x \to 2 lim (\frac{1}{x ^{2}})

x \to π lim (\frac{x}{3} - \frac{2 π}{3})

\frac{d ^{2} x}{d t ^{2}} + 4 x = t^{2}

((5 - 120 x)^{2} + (150 x)^{2})

What is the integral of 5/(x^2-1) ?
The integral of 5/(x^2-1) is -5(1/2 ln|x+1|-1/2 ln|x-1|)+C