What is the integral of (10)/((x-2)(x+8)) ?

The integral of (10)/((x-2)(x+8)) is ln|x-2|-ln|x+8|+C

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integral of (10)/((x-2)(x+8))

\int \frac{10}{( x - 2 ) ( x + 8 )} d x

ln ∣ x - 2 ∣ - ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{10}{( x - 2 ) ( x + 8 )} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{( x - 2 ) ( x + 8 )} d x

Take the partial fraction of

\frac{1}{( x - 2 ) ( x + 8 )} : \frac{1}{10 ( x - 2 )} - \frac{1}{10 ( x + 8 )}

= 10 \cdot \int \frac{1}{10 ( x - 2 )} - \frac{1}{10 ( x + 8 )} 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

= 10 (\int \frac{1}{10 ( x - 2 )} d x - \int \frac{1}{10 ( x + 8 )} d x)

\int \frac{1}{10 ( x - 2 )} d x = \frac{1}{10} ln ∣ x - 2 ∣

\int \frac{1}{10 ( x + 8 )} d x = \frac{1}{10} ln ∣ x + 8 ∣

= 10 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x + 8 ∣)

Simplify

10 (\frac{1}{10} ln ∣ x - 2 ∣ - \frac{1}{10} ln ∣ x + 8 ∣) : ln ∣ x - 2 ∣ - ln ∣ x + 8 ∣

= ln ∣ x - 2 ∣ - ln ∣ x + 8 ∣

Add a constant to the solution

= ln ∣ x - 2 ∣ - ln ∣ x + 8 ∣ + C

y^{^{''}} = e^{x^{2}} + 11 x

\int_{0}^{100} e^{- 0.01 z} d z

e^{- 3 (t - 3)}

x \to 6 lim (8 x + ∣ x - 6 ∣)

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

What is the integral of (10)/((x-2)(x+8)) ?
The integral of (10)/((x-2)(x+8)) is ln|x-2|-ln|x+8|+C