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

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

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integral of 1/(4-x^2)

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

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

Solution steps

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

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{x}{2}

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

Simplify

\frac{1}{2} (\frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2}) : \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

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

Add a constant to the solution

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

\frac{d y}{d x} = (- \frac{3 y}{3 x + 2 y})

- 6 y^{^{''}} + 24 y = 0

\frac{\partial}{\partial y} (x y e^{x z})

x \to - \infty lim (x^{2} + 2 x)

\frac{d ^{2} x}{d x ^{2}}

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