What is the integral of (7x-1)/(x^2+x-2) ?

The integral of (7x-1)/(x^2+x-2) is

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integral of (7x-1)/(x^2+x-2)

\int \frac{7 x - 1}{x ^{2} + x - 2} d x

7 (\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{7 x - 1}{x ^{2} + x - 2} d x

Expand

\frac{7 x - 1}{x ^{2} + x - 2} : \frac{7 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{7 x}{x ^{2} + x - 2} d x - \int \frac{1}{x ^{2} + x - 2} d x

\int \frac{7 x}{x ^{2} + x - 2} d x = 7 (\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)

= 7 (\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))

Simplify

7 (\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)) : 7 (\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)

= 7 (\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

= 7 (\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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