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))

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

\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

y^{^{'}} + 3 y = x e^{- 2 x} + e^{- 3 x}

y^{^{''}} = x (y^{^{'}})^{2}

\frac{x}{1 + x ^{2}}, at x = 0

\int 36 x e^{- 9 x} d x

\frac{2 x + 1}{x + 5} \cdot (3 x - 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