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

The integral of (x^2)/((1-x^2)^2) is -1/4 ln|x+1|-1/(4(x+1))+1/4 ln|x-1|-1/(4(x-1))+C

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

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

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

Solution steps

\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}}

= \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

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

Add a constant to the solution

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

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

\frac{d}{d x} (7 x - 4 x)

f (t) = cos (t)

x \to 0 lim (2 x cos (\frac{1}{x}) - 2)

f (x) = (4 x + 2)^{- 1}

What is the integral of (x^2)/((1-x^2)^2) ?
The integral of (x^2)/((1-x^2)^2) is -1/4 ln|x+1|-1/(4(x+1))+1/4 ln|x-1|-1/(4(x-1))+C