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

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

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

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

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

Solution steps

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

Expand

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

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

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

Add a constant to the solution

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

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