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

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

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integral of 1/((1-x)(1+x))

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

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

Solution steps

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

Expand

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

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

Use the common integral:

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

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

Add a constant to the solution

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

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