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

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

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

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

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

Solution steps

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

Expand

\frac{5 x + 1}{x ^{2} + x - 2} : \frac{5 x}{x ^{2} + x - 2} + \frac{1}{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{5 x}{x ^{2} + x - 2} d x + \int \frac{1}{x ^{2} + x - 2} d x

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

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

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

Add a constant to the solution

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

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