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

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

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

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

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

Solution steps

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

Expand

\frac{1 - x}{1 - x ^{2}} : \frac{1}{1 - x ^{2}} - \frac{x}{1 - x ^{2}}

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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