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

x y^{2} \frac{d y}{d x} = y^{3} + x^{3}, y (1) = 5

\frac{\partial}{\partial x} (\frac{2 x}{x - y})

5 t \cdot \frac{d y}{d t} + 2 y = t

y^{\frac{3}{2}} (y + c e^{y})

\frac{\partial}{\partial x} (\frac{a \cdot b}{c \cdot x})

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