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

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

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

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

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

Solution steps

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

Expand

(- x + 1) (x + 1) : 1 - x^{2}

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int_{0}^{1} \frac{37}{4 y - 1} d y

\int_{0}^{3} arctan (x) d x

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

\int cos (x) + x cos (y) - y d y

\frac{\partial}{\partial x} (\frac{3 y}{2 x})

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