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

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

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integral of 1/(1-u^2)

\int \frac{1}{1 - u ^{2}} d u

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

Solution steps

\int \frac{1}{1 - u ^{2}} d u

Use the common integral:

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

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

Add a constant to the solution

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

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

t \frac{d y}{d t} - y = y t

\frac{\partial}{\partial x} (e^{- 7 x} cos (7 π t))

n = 1 \sum \infty \frac{n ^{2}}{n + 5}

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