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

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

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integral of 4/(x^2-16)

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

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

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 4 \cdot \int \frac{1}{4 ( u ^{2} - 1 )} d u

Take the constant out:

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

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

Factor

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

= 4 \cdot \frac{1}{4} \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

= 4 \cdot \frac{1}{4} (- \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}

= 4 \cdot \frac{1}{4} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x + 3 x y^{2} + x^{2} y)

\frac{1}{s ( s + 1 ) ^{3}}

\frac{x ^{3}}{x - 1}

\int_{0}^{1} \frac{21}{x ^{5}} d x

y^{^{''}} + y^{^{'}} + 5 y = 0

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