What is the integral of 8/(x^2-4) ?

The integral of 8/(x^2-4) is -4(1/2 ln| x/2+1|-1/2 ln| x/2-1|)+C

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integral of 8/(x^2-4)

\int \frac{8}{x ^{2} - 4} d x

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

Solution steps

\int \frac{8}{x ^{2} - 4} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2} - 4} d x

Apply Integral Substitution

= 8 \cdot \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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} (- \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}

= 8 \cdot \frac{1}{2} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{2}

= 8 \cdot \frac{1}{2} (- (\frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2}))

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} = 3 y

y = 0, y = x^{2} - x

\int \frac{sec ( x ) + cos ( x )}{2 cos ( x )} d x

\int \frac{5}{x - 4} d x

3 x \frac{d y}{d x} + y = 0, y (1) = 1

What is the integral of 8/(x^2-4) ?
The integral of 8/(x^2-4) is -4(1/2 ln| x/2+1|-1/2 ln| x/2-1|)+C