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

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

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

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

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

Solution steps

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

Use the common integral:

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

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

Add a constant to the solution

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

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\frac{d y}{d x} = 3 - y

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