What is the integral of (x+2)/((x-1)(x+3)(x-5)) ?

The integral of (x+2)/((x-1)(x+3)(x-5)) is -3/16 ln|x-1|-1/32 ln|x+3|+7/32 ln|x-5|+C

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integral of (x+2)/((x-1)(x+3)(x-5))

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

- \frac{3}{16} ln ∣ x - 1 ∣ - \frac{1}{32} ln ∣ x + 3 ∣ + \frac{7}{32} ln ∣ x - 5 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x + 2}{( x - 1 ) ( x + 3 ) ( x - 5 )} : - \frac{3}{16 ( x - 1 )} - \frac{1}{32 ( x + 3 )} + \frac{7}{32 ( x - 5 )}

= \int - \frac{3}{16 ( x - 1 )} - \frac{1}{32 ( x + 3 )} + \frac{7}{32 ( x - 5 )} 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{3}{16 ( x - 1 )} d x - \int \frac{1}{32 ( x + 3 )} d x + \int \frac{7}{32 ( x - 5 )} d x

\int \frac{3}{16 ( x - 1 )} d x = \frac{3}{16} ln ∣ x - 1 ∣

\int \frac{1}{32 ( x + 3 )} d x = \frac{1}{32} ln ∣ x + 3 ∣

\int \frac{7}{32 ( x - 5 )} d x = \frac{7}{32} ln ∣ x - 5 ∣

= - \frac{3}{16} ln ∣ x - 1 ∣ - \frac{1}{32} ln ∣ x + 3 ∣ + \frac{7}{32} ln ∣ x - 5 ∣

Add a constant to the solution

= - \frac{3}{16} ln ∣ x - 1 ∣ - \frac{1}{32} ln ∣ x + 3 ∣ + \frac{7}{32} ln ∣ x - 5 ∣ + C

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What is the integral of (x+2)/((x-1)(x+3)(x-5)) ?
The integral of (x+2)/((x-1)(x+3)(x-5)) is -3/16 ln|x-1|-1/32 ln|x+3|+7/32 ln|x-5|+C