What is the integral of 6/((x-2)(x+4)) ?

The integral of 6/((x-2)(x+4)) is ln|x-2|-ln|x+4|+C

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integral of 6/((x-2)(x+4))

\int \frac{6}{( x - 2 ) ( x + 4 )} d x

ln ∣ x - 2 ∣ - ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{6}{( x - 2 ) ( x + 4 )} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( x - 2 ) ( x + 4 )} d x

Take the partial fraction of

\frac{1}{( x - 2 ) ( x + 4 )} : \frac{1}{6 ( x - 2 )} - \frac{1}{6 ( x + 4 )}

= 6 \cdot \int \frac{1}{6 ( x - 2 )} - \frac{1}{6 ( x + 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int \frac{1}{6 ( x - 2 )} d x - \int \frac{1}{6 ( x + 4 )} d x)

\int \frac{1}{6 ( x - 2 )} d x = \frac{1}{6} ln ∣ x - 2 ∣

\int \frac{1}{6 ( x + 4 )} d x = \frac{1}{6} ln ∣ x + 4 ∣

= 6 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣)

Simplify

6 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣) : ln ∣ x - 2 ∣ - ln ∣ x + 4 ∣

= ln ∣ x - 2 ∣ - ln ∣ x + 4 ∣

Add a constant to the solution

= ln ∣ x - 2 ∣ - ln ∣ x + 4 ∣ + C

\frac{\partial}{\partial x} ((3 x + 4 y)^{5})

y = \frac{x ^{5}}{1 - x ^{4}}

x \to \infty lim (sin (0))

n = 2 \sum \infty \frac{ln ( n )}{7 n}

n = 0 \sum \infty \frac{2 n}{n ^{2} + 1}

What is the integral of 6/((x-2)(x+4)) ?
The integral of 6/((x-2)(x+4)) is ln|x-2|-ln|x+4|+C