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

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

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

Take the partial fraction of

\frac{1}{( 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

f (x) = \frac{5}{x - 2}, at x = 4

\int tan^{2} (7 x) d x

\int_{1}^{2} 12 x^{2} 4^{x^{3}} d x

- 5 sin (2 t)

e^{- x + 1} - e^{- x + 1} x

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