What is the integral of 8/((x-2)(x+6)) ?

The integral of 8/((x-2)(x+6)) is ln|x-2|-ln|x+6|+C

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integral of 8/((x-2)(x+6))

\int \frac{8}{( x - 2 ) ( x + 6 )} d x

ln ∣ x - 2 ∣ - ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{8}{( x - 2 ) ( x + 6 )} 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 ) ( x + 6 )} d x

Take the partial fraction of

\frac{1}{( x - 2 ) ( x + 6 )} : \frac{1}{8 ( x - 2 )} - \frac{1}{8 ( x + 6 )}

= 8 \cdot \int \frac{1}{8 ( x - 2 )} - \frac{1}{8 ( x + 6 )} 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

= 8 (\int \frac{1}{8 ( x - 2 )} d x - \int \frac{1}{8 ( x + 6 )} d x)

\int \frac{1}{8 ( x - 2 )} d x = \frac{1}{8} ln ∣ x - 2 ∣

\int \frac{1}{8 ( x + 6 )} d x = \frac{1}{8} ln ∣ x + 6 ∣

= 8 (\frac{1}{8} ln ∣ x - 2 ∣ - \frac{1}{8} ln ∣ x + 6 ∣)

Simplify

8 (\frac{1}{8} ln ∣ x - 2 ∣ - \frac{1}{8} ln ∣ x + 6 ∣) : ln ∣ x - 2 ∣ - ln ∣ x + 6 ∣

= ln ∣ x - 2 ∣ - ln ∣ x + 6 ∣

Add a constant to the solution

= ln ∣ x - 2 ∣ - ln ∣ x + 6 ∣ + C

y = 9 x^{2} - x^{e}

\frac{d}{d x} (- C x - d)

\frac{\partial}{\partial x} (x^{2} y^{2} e^{6 x y})

\int \frac{3 - 18 x}{16 - 9 x ^{2}} d x

y^{^{''}} + y^{^{'}} - 2 y = e^{- 2 t}

What is the integral of 8/((x-2)(x+6)) ?
The integral of 8/((x-2)(x+6)) is ln|x-2|-ln|x+6|+C