What is the integral of (6x+8)/((x+3)(x+5)(x-2)) ?

The integral of (6x+8)/((x+3)(x+5)(x-2)) is ln|x+3|-11/7 ln|x+5|+4/7 ln|x-2|+C

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integral of (6x+8)/((x+3)(x+5)(x-2))

\int \frac{6 x + 8}{( x + 3 ) ( x + 5 ) ( x - 2 )} d x

ln ∣ x + 3 ∣ - \frac{11}{7} ln ∣ x + 5 ∣ + \frac{4}{7} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{6 x + 8}{( x + 3 ) ( x + 5 ) ( x - 2 )} d x

Take the partial fraction of

\frac{6 x + 8}{( x + 3 ) ( x + 5 ) ( x - 2 )} : \frac{1}{x + 3} - \frac{11}{7 ( x + 5 )} + \frac{4}{7 ( x - 2 )}

= \int \frac{1}{x + 3} - \frac{11}{7 ( x + 5 )} + \frac{4}{7 ( 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

= \int \frac{1}{x + 3} d x - \int \frac{11}{7 ( x + 5 )} d x + \int \frac{4}{7 ( x - 2 )} d x

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

\int \frac{11}{7 ( x + 5 )} d x = \frac{11}{7} ln ∣ x + 5 ∣

\int \frac{4}{7 ( x - 2 )} d x = \frac{4}{7} ln ∣ x - 2 ∣

= ln ∣ x + 3 ∣ - \frac{11}{7} ln ∣ x + 5 ∣ + \frac{4}{7} ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x + 3 ∣ - \frac{11}{7} ln ∣ x + 5 ∣ + \frac{4}{7} ln ∣ x - 2 ∣ + C

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What is the integral of (6x+8)/((x+3)(x+5)(x-2)) ?
The integral of (6x+8)/((x+3)(x+5)(x-2)) is ln|x+3|-11/7 ln|x+5|+4/7 ln|x-2|+C