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

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

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

integral

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{x}{( x ^{2} + 9 )}

x (x + 4)^{8}

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

ln^{5} (x^{2})

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