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

The integral of-(x^2)/((-x+1)(x+1)) is -1/2 ln|x+1|+1/2 ln|x-1|+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{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + x + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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)

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (A e^{x} + B e^{- x})

x y^{^{'}} = ((1 - x) y + x^{2} + x^{3}), y (3) = 2

x \to 3 - lim (\frac{x}{∣ x ∣})

\frac{\partial}{\partial x} (2 y^{7} + 7 y^{9} x^{6})

\int 2 x (x + 1) d x

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