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

\frac{1}{x - 9}

\frac{d}{d x} (\frac{1 - 6 x}{5 + x})

\int \frac{1}{8 - x} d x

\frac{d}{d x} (7 x + 8)

z e^{z}

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