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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

\int \frac{1}{2 ( x - 1 )} d x = \frac{1}{2} ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x + 1 )}{3})

\frac{d}{d x} (\frac{x ^{2}}{sin ( a ) x})

\int \frac{1}{x ^{2} + x ^{4}} d x

y^{^{'}} - y = 2 cos (5 t)

x \to 2 lim (- x^{3})

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