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

The integral of (3x+2)/((x+5)(x-2)) is

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

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

6 (\frac{1}{4} ln 4 x^{2} + 12 x - 40 + \frac{3}{28} (ln \frac{10}{7} + \frac{2}{7} x - ln - \frac{4}{7} + \frac{2}{7} x)) + 2 (- \frac{1}{7} ln ∣ x + 5 ∣ + \frac{1}{7} ln ∣ x - 2 ∣) + C

Solution steps

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

Expand

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

\int \frac{3 x}{( x + 5 ) ( x - 2 )} d x = 6 (\frac{1}{4} ln 4 x^{2} + 12 x - 40 + \frac{3}{28} (ln \frac{10}{7} + \frac{2}{7} x - ln - \frac{4}{7} + \frac{2}{7} x))

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

= 6 (\frac{1}{4} ln 4 x^{2} + 12 x - 40 + \frac{3}{28} (ln \frac{10}{7} + \frac{2}{7} x - ln - \frac{4}{7} + \frac{2}{7} x)) + 2 (- \frac{1}{7} ln ∣ x + 5 ∣ + \frac{1}{7} ln ∣ x - 2 ∣)

Add a constant to the solution

= 6 (\frac{1}{4} ln 4 x^{2} + 12 x - 40 + \frac{3}{28} (ln \frac{10}{7} + \frac{2}{7} x - ln - \frac{4}{7} + \frac{2}{7} x)) + 2 (- \frac{1}{7} ln ∣ x + 5 ∣ + \frac{1}{7} ln ∣ x - 2 ∣) + C

s \to - 8 lim (4 (s) - 2)

\frac{d}{d x} (\frac{2}{x ^{2}} - 5 x)

\frac{d}{d x} (- \frac{9}{( 6 x + 3 ) ^{\frac{3}{2}}})

g (x) = x^{2} (1 - 2 x)

\frac{\partial}{\partial x} (x - y)

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