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

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

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

integral

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{1}{2} - 3 cos^{2} (x)

\frac{1}{( x + x ^{2} )}

\frac{2}{t}

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

\frac{1}{( x + x ^{2} )}

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