What is the integral of (16)/(x^2-4) ?

The integral of (16)/(x^2-4) is -8(1/2 ln| x/2+1|-1/2 ln| x/2-1|)+C

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integral of (16)/(x^2-4)

\int \frac{16}{x ^{2} - 4} d x

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

Solution steps

\int \frac{16}{x ^{2} - 4} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{x ^{2} - 4} d x

Apply Integral Substitution

= 16 \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

= 16 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 16 \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

= 16 \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}

= 16 \cdot \frac{1}{2} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{2}

= 16 \cdot \frac{1}{2} (- (\frac{ln \frac{x}{2} + 1}{2} - \frac{ln \frac{x}{2} - 1}{2}))

Simplify

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

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

Add a constant to the solution

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

- 10 x + 11, x^{2} - 5 x - 13

y = \frac{5 x}{( x + 1 ) ^{2}}, (0, 0)

x y^{^{'}} + y^{2} = 0

\int - x d x

\frac{\partial}{\partial x} (y \cdot arctan (\frac{x}{z}))

What is the integral of (16)/(x^2-4) ?
The integral of (16)/(x^2-4) is -8(1/2 ln| x/2+1|-1/2 ln| x/2-1|)+C