What is the integral of (8x-1)/((x-2)(x+3)) ?

The integral of (8x-1)/((x-2)(x+3)) is

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integral of (8x-1)/((x-2)(x+3))

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

4 ln 4 x^{2} + 4 x - 24 + \frac{4}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) - \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln ∣ x + 3 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{8 x}{( x - 2 ) ( x + 3 )} d x = 16 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

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

= 16 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - (\frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{5} ln ∣ x + 3 ∣)

Simplify

16 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - (\frac{1}{5} ln ∣ x - 2 ∣ - \frac{1}{5} ln ∣ x + 3 ∣) : 4 ln 4 x^{2} + 4 x - 24 + \frac{4}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) - \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln ∣ x + 3 ∣

= 4 ln 4 x^{2} + 4 x - 24 + \frac{4}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) - \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln ∣ x + 3 ∣

Add a constant to the solution

= 4 ln 4 x^{2} + 4 x - 24 + \frac{4}{5} (ln \frac{2}{5} x + \frac{6}{5} - ln \frac{2}{5} x - \frac{4}{5}) - \frac{1}{5} ln ∣ x - 2 ∣ + \frac{1}{5} ln ∣ x + 3 ∣ + C

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What is the integral of (8x-1)/((x-2)(x+3)) ?
The integral of (8x-1)/((x-2)(x+3)) is