What is the integral of (e^{-ln(1-x^2)})/(x^2) ?

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

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integral of (e^{-ln(1-x^2)})/(x^2)

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

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

Solution steps

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

Simplify

\frac{e ^{- l n (1 - x^{2})}}{x ^{2}} : \frac{1}{x ^{2} ( 1 - x ^{2} )}

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

\frac{2 x ^{2} + 6 x + 6}{x}

\int x 8 - x^{2} d x

(- 1, 19), (5, - 41)

tan (ln (x))

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