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

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

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

integral

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

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

Solution steps

\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

in t e g r a l \frac{x ^{3}}{( x + 1 ) ^{2}}

in t e g r a l (x - 1) (x + 3)^{3}

in t e g r a l (e^{y})^{2}

in t e g r a l \frac{1}{( x ^{3} + x ^{2} + x )}

in t e g r a l \frac{1}{( x ^{2} - 1 )}

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