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

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

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

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

- x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

\frac{- x ^{2} - 1}{x ^{2} - 1} : - \frac{x ^{2}}{x ^{2} - 1} - \frac{1}{x ^{2} - 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

= - x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= - x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

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