What is the integral of 1/(4-y^2) ?

The integral of 1/(4-y^2) is 1/4 (ln| y/2+1|-ln| y/2-1|)+C

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integral of 1/(4-y^2)

\int \frac{1}{4 - y ^{2}} d y

\frac{1}{4} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1) + C

Solution steps

\int \frac{1}{4 - y ^{2}} d y

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} 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}{4} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1)

= \frac{1}{4} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1)

Add a constant to the solution

= \frac{1}{4} (ln \frac{y}{2} + 1 - ln \frac{y}{2} - 1) + C

\int sin^{8} (3 x) cos^{5} (3 x) d x

\frac{d}{d x} (10 x^{4})

f (x) = ln (x^{2} - 8 x + 16)

\frac{\partial}{\partial x} (cos^{2} (θ))

\frac{d}{d x} (x^{(\frac{2}{3})})

What is the integral of 1/(4-y^2) ?
The integral of 1/(4-y^2) is 1/4 (ln| y/2+1|-ln| y/2-1|)+C