What is the integral of 8/((x-5)(x+3)) ?

The integral of 8/((x-5)(x+3)) is ln|x-5|-ln|x+3|+C

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integral of 8/((x-5)(x+3))

\int \frac{8}{( x - 5 ) ( x + 3 )} d x

ln ∣ x - 5 ∣ - ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{8}{( x - 5 ) ( x + 3 )} 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 - 5 ) ( x + 3 )} d x

Take the partial fraction of

\frac{1}{( x - 5 ) ( x + 3 )} : \frac{1}{8 ( x - 5 )} - \frac{1}{8 ( x + 3 )}

= 8 \cdot \int \frac{1}{8 ( x - 5 )} - \frac{1}{8 ( x + 3 )} 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 - 5 )} d x - \int \frac{1}{8 ( x + 3 )} d x)

\int \frac{1}{8 ( x - 5 )} d x = \frac{1}{8} ln ∣ x - 5 ∣

\int \frac{1}{8 ( x + 3 )} d x = \frac{1}{8} ln ∣ x + 3 ∣

= 8 (\frac{1}{8} ln ∣ x - 5 ∣ - \frac{1}{8} ln ∣ x + 3 ∣)

Simplify

8 (\frac{1}{8} ln ∣ x - 5 ∣ - \frac{1}{8} ln ∣ x + 3 ∣) : ln ∣ x - 5 ∣ - ln ∣ x + 3 ∣

= ln ∣ x - 5 ∣ - ln ∣ x + 3 ∣

Add a constant to the solution

= ln ∣ x - 5 ∣ - ln ∣ x + 3 ∣ + C

\int e^{x} \cdot ln (x) d x

n = 1 \sum \infty \frac{n}{e ^{n^{2}}}

\frac{\partial}{\partial y} (e^{x y} - (x + 1) cos (y))

x \to 0 lim (\frac{x ^{3} + 7 x}{x})

y = x^{2} - 5, y = 4

What is the integral of 8/((x-5)(x+3)) ?
The integral of 8/((x-5)(x+3)) is ln|x-5|-ln|x+3|+C