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

The integral of (-x^2)/(2(1-x^2)^2) is -(x^2ln|x-1|-x^2ln|x+1|-2x+ln|x+1|-ln|x-1|)/(8(x+1)(x-1))+C

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

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

- \frac{x ^{2} ln ∣ x - 1 ∣ - x ^{2} ln ∣ x + 1 ∣ - 2 x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣}{8 ( x + 1 ) ( x - 1 )} + C

Solution steps

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

Take the constant out:

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

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

Simplify

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

Take the partial fraction of

\frac{x ^{2}}{( 1 - x ^{2} ) ^{2}} : - \frac{1}{4 ( x + 1 )} + \frac{1}{4 ( x + 1 ) ^{2}} + \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 1 ) ^{2}}

= (- \frac{1}{2}) \cdot \int - \frac{1}{4 ( x + 1 )} + \frac{1}{4 ( x + 1 ) ^{2}} + \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 1 ) ^{2}} 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

= (- \frac{1}{2}) (- \int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{4 ( x + 1 ) ^{2}} d x + \int \frac{1}{4 ( x - 1 )} d x + \int \frac{1}{4 ( x - 1 ) ^{2}} d x)

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x + 1 ) ^{2}} d x = - \frac{1}{4 ( x + 1 )}

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{1}{4 ( x - 1 ) ^{2}} d x = - \frac{1}{4 ( x - 1 )}

= (- \frac{1}{2}) (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )})

Simplify

(- \frac{1}{2}) (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) : - \frac{x ^{2} ln ∣ x - 1 ∣ - x ^{2} ln ∣ x + 1 ∣ - 2 x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣}{8 ( x + 1 ) ( x - 1 )}

= - \frac{x ^{2} ln ∣ x - 1 ∣ - x ^{2} ln ∣ x + 1 ∣ - 2 x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣}{8 ( x + 1 ) ( x - 1 )}

Add a constant to the solution

= - \frac{x ^{2} ln ∣ x - 1 ∣ - x ^{2} ln ∣ x + 1 ∣ - 2 x + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣}{8 ( x + 1 ) ( x - 1 )} + C

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

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\int (x + 2)^{2} \cdot ln (5 x) d x

What is the integral of (-x^2)/(2(1-x^2)^2) ?
The integral of (-x^2)/(2(1-x^2)^2) is -(x^2ln|x-1|-x^2ln|x+1|-2x+ln|x+1|-ln|x-1|)/(8(x+1)(x-1))+C