What is the integral of (17x+4)/(3x^2-7x-6) ?

The integral of (17x+4)/(3x^2-7x-6) is

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integral of (17x+4)/(3x^2-7x-6)

\int \frac{17 x + 4}{3 x ^{2} - 7 x - 6} d x

17 (\frac{2}{33} ln ∣ 3 x + 2 ∣ + \frac{3}{11} ln ∣ x - 3 ∣) - \frac{4}{11} (ln \frac{6}{11} x + \frac{4}{11} - ln \frac{6}{11} x - \frac{18}{11}) + C

Solution steps

\int \frac{17 x + 4}{3 x ^{2} - 7 x - 6} d x

Expand

\frac{17 x + 4}{3 x ^{2} - 7 x - 6} : \frac{17 x}{3 x ^{2} - 7 x - 6} + \frac{4}{3 x ^{2} - 7 x - 6}

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{17 x}{3 x ^{2} - 7 x - 6} d x + \int \frac{4}{3 x ^{2} - 7 x - 6} d x

\int \frac{17 x}{3 x ^{2} - 7 x - 6} d x = 17 (\frac{2}{33} ln ∣ 3 x + 2 ∣ + \frac{3}{11} ln ∣ x - 3 ∣)

\int \frac{4}{3 x ^{2} - 7 x - 6} d x = - \frac{4}{11} (ln \frac{6}{11} x + \frac{4}{11} - ln \frac{6}{11} x - \frac{18}{11})

= 17 (\frac{2}{33} ln ∣ 3 x + 2 ∣ + \frac{3}{11} ln ∣ x - 3 ∣) - \frac{4}{11} (ln \frac{6}{11} x + \frac{4}{11} - ln \frac{6}{11} x - \frac{18}{11})

Add a constant to the solution

= 17 (\frac{2}{33} ln ∣ 3 x + 2 ∣ + \frac{3}{11} ln ∣ x - 3 ∣) - \frac{4}{11} (ln \frac{6}{11} x + \frac{4}{11} - ln \frac{6}{11} x - \frac{18}{11}) + C

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What is the integral of (17x+4)/(3x^2-7x-6) ?
The integral of (17x+4)/(3x^2-7x-6) is