What is the integral of (x-3)/(2x^3-8x) ?

The integral of (x-3)/(2x^3-8x) is 3/8 ln|x|-5/16 ln|x+2|-1/16 ln|x-2|+C

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integral of (x-3)/(2x^3-8x)

\int \frac{x - 3}{2 x ^{3} - 8 x} d x

\frac{3}{8} ln ∣ x ∣ - \frac{5}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x - 3}{2 x ^{3} - 8 x} d x

Take the partial fraction of

\frac{x - 3}{2 x ^{3} - 8 x} : \frac{3}{8 x} - \frac{5}{16 ( x + 2 )} - \frac{1}{16 ( x - 2 )}

= \int \frac{3}{8 x} - \frac{5}{16 ( x + 2 )} - \frac{1}{16 ( 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{3}{8 x} d x - \int \frac{5}{16 ( x + 2 )} d x - \int \frac{1}{16 ( x - 2 )} d x

\int \frac{3}{8 x} d x = \frac{3}{8} ln ∣ x ∣

\int \frac{5}{16 ( x + 2 )} d x = \frac{5}{16} ln ∣ x + 2 ∣

\int \frac{1}{16 ( x - 2 )} d x = \frac{1}{16} ln ∣ x - 2 ∣

= \frac{3}{8} ln ∣ x ∣ - \frac{5}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{3}{8} ln ∣ x ∣ - \frac{5}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x - 2 ∣ + C

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What is the integral of (x-3)/(2x^3-8x) ?
The integral of (x-3)/(2x^3-8x) is 3/8 ln|x|-5/16 ln|x+2|-1/16 ln|x-2|+C