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

9 e^{x} (cos (x) - sin (x))

n = 0 \sum \infty (- x)^{n}

x \to 0 + lim (\frac{x}{ln ( \frac{1}{x} )})

\frac{\partial}{\partial x} (\frac{y}{2 x})

x \cdot sin (\frac{4}{x})

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