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

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

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

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

- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x + C

Solution steps

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

\frac{x ^{2} + 1}{x ^{2} - 1} = \frac{2}{x ^{2} - 1} + 1

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

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

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

\int 1 d x = x

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x

Add a constant to the solution

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

\frac{\partial}{\partial x} (- \frac{3}{x ^{2}})

\frac{d y}{d x}, y = x^{- 10 c o s (x)}

f (x) = \frac{x ^{2} - 1}{x ^{2} + x + 1}, (1, 0)

(1 + x) \frac{d y}{d x} - x y = x + x^{2}

\frac{d}{d x} (x^{8 y})

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