What is the integral of 5/((x-3)(x+2)) ?

The integral of 5/((x-3)(x+2)) is ln|x-3|-ln|x+2|+C

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integral of 5/((x-3)(x+2))

\int \frac{5}{( x - 3 ) ( x + 2 )} d x

ln ∣ x - 3 ∣ - ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{5}{( x - 3 ) ( x + 2 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{( x - 3 ) ( x + 2 )} d x

Take the partial fraction of

\frac{1}{( x - 3 ) ( x + 2 )} : \frac{1}{5 ( x - 3 )} - \frac{1}{5 ( x + 2 )}

= 5 \cdot \int \frac{1}{5 ( x - 3 )} - \frac{1}{5 ( x + 2 )} 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

= 5 (\int \frac{1}{5 ( x - 3 )} d x - \int \frac{1}{5 ( x + 2 )} d x)

\int \frac{1}{5 ( x - 3 )} d x = \frac{1}{5} ln ∣ x - 3 ∣

\int \frac{1}{5 ( x + 2 )} d x = \frac{1}{5} ln ∣ x + 2 ∣

= 5 (\frac{1}{5} ln ∣ x - 3 ∣ - \frac{1}{5} ln ∣ x + 2 ∣)

Simplify

5 (\frac{1}{5} ln ∣ x - 3 ∣ - \frac{1}{5} ln ∣ x + 2 ∣) : ln ∣ x - 3 ∣ - ln ∣ x + 2 ∣

= ln ∣ x - 3 ∣ - ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ x - 3 ∣ - ln ∣ x + 2 ∣ + C

\frac{d}{d x} (\frac{4 x}{1 - cot ( x )})

\frac{2}{4 - x}

\frac{d}{d x} (x - π)

y = \frac{1}{x ^{4}}, (- 2, \frac{1}{16})

\int (2 e^{x}) d x

What is the integral of 5/((x-3)(x+2)) ?
The integral of 5/((x-3)(x+2)) is ln|x-3|-ln|x+2|+C