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

The integral of (2x+1)/((x+5)(x-2)) is

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

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

ln 4 x^{2} + 12 x - 40 + \frac{3}{7} (ln \frac{2}{7} x + \frac{10}{7} - ln \frac{2}{7} x - \frac{4}{7}) - \frac{1}{7} ln ∣ x + 5 ∣ + \frac{1}{7} ln ∣ x - 2 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{2 x}{( x + 5 ) ( x - 2 )} d x = ln 4 x^{2} + 12 x - 40 + \frac{3}{7} (ln \frac{2}{7} x + \frac{10}{7} - ln \frac{2}{7} x - \frac{4}{7})

\int \frac{1}{( x + 5 ) ( x - 2 )} d x = - \frac{1}{7} ln ∣ x + 5 ∣ + \frac{1}{7} ln ∣ x - 2 ∣

= ln 4 x^{2} + 12 x - 40 + \frac{3}{7} (ln \frac{2}{7} x + \frac{10}{7} - ln \frac{2}{7} x - \frac{4}{7}) - \frac{1}{7} ln ∣ x + 5 ∣ + \frac{1}{7} ln ∣ x - 2 ∣

Add a constant to the solution

= ln 4 x^{2} + 12 x - 40 + \frac{3}{7} (ln \frac{2}{7} x + \frac{10}{7} - ln \frac{2}{7} x - \frac{4}{7}) - \frac{1}{7} ln ∣ x + 5 ∣ + \frac{1}{7} ln ∣ x - 2 ∣ + C

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