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

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

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

integral

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

\frac{1}{2 2} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Substitute back

u = \frac{y}{2}

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

Simplify

\frac{1}{2} \frac{ln \frac{y}{2} + 1}{2} - \frac{ln \frac{y}{2} - 1}{2} : \frac{1}{2 2} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1)

= \frac{1}{2 2} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1)

Add a constant to the solution

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

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