What is the integral of (4x+5)/(x^2+x-2) ?

The integral of (4x+5)/(x^2+x-2) is

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integral of (4x+5)/(x^2+x-2)

\int \frac{4 x + 5}{x ^{2} + x - 2} d x

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

Solution steps

\int \frac{4 x + 5}{x ^{2} + x - 2} d x

Expand

\frac{4 x + 5}{x ^{2} + x - 2} : \frac{4 x}{x ^{2} + x - 2} + \frac{5}{x ^{2} + 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{4 x}{x ^{2} + x - 2} d x + \int \frac{5}{x ^{2} + x - 2} d x

\int \frac{4 x}{x ^{2} + x - 2} d x = 4 (\frac{1}{3} ln ∣ x - 1 ∣ + \frac{2}{3} ln ∣ x + 2 ∣)

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

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

Add a constant to the solution

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