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

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

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integral of-1/(x^2-1)

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

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

Solution steps

\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

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

Factor

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

= - \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

= - (- \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}

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{5} t ln (t) d t

y^{^{'}} = sec^{2} (x) (2 + y)^{2}, y (π) = - 5

y^{^{''}} + 4 y = 3 sin (2 x)

2^{x^{2} + 1}

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