What is the integral of (4x^3-x)/(x^2-x-30) ?

The integral of (4x^3-x)/(x^2-x-30) is

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integral of (4x^3-x)/(x^2-x-30)

\int \frac{4 x ^{3} - x}{x ^{2} - x - 30} d x

2 x^{2} + 4 x + \frac{123}{2} ln 4 x^{2} - 4 x - 120 - \frac{120}{11} ln ∣ x + 5 ∣ + \frac{120}{11} ln ∣ x - 6 ∣ - \frac{62}{11} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11}) + \frac{1}{22} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11}) + C

Solution steps

\int \frac{4 x ^{3} - x}{x ^{2} - x - 30} d x

Expand

\frac{4 x ^{3} - x}{x ^{2} - x - 30} : \frac{4 x ^{3}}{x ^{2} - x - 30} - \frac{x}{x ^{2} - x - 30}

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 ^{3}}{x ^{2} - x - 30} d x - \int \frac{x}{x ^{2} - x - 30} d x

\int \frac{4 x ^{3}}{x ^{2} - x - 30} d x = 4 (\frac{x ^{2}}{2} + x + 62 (\frac{1}{4} ln 4 x^{2} - 4 x - 120 - \frac{1}{44} (ln \frac{10}{11} + \frac{2}{11} x - ln - \frac{12}{11} + \frac{2}{11} x)) + 30 (- \frac{1}{11} ln ∣ x + 5 ∣ + \frac{1}{11} ln ∣ x - 6 ∣))

\int \frac{x}{x ^{2} - x - 30} d x = 2 (\frac{1}{4} ln 4 x^{2} - 4 x - 120 - \frac{1}{44} (ln \frac{10}{11} + \frac{2}{11} x - ln - \frac{12}{11} + \frac{2}{11} x))

= 4 (\frac{x ^{2}}{2} + x + 62 (\frac{1}{4} ln 4 x^{2} - 4 x - 120 - \frac{1}{44} (ln \frac{10}{11} + \frac{2}{11} x - ln - \frac{12}{11} + \frac{2}{11} x)) + 30 (- \frac{1}{11} ln ∣ x + 5 ∣ + \frac{1}{11} ln ∣ x - 6 ∣)) - 2 (\frac{1}{4} ln 4 x^{2} - 4 x - 120 - \frac{1}{44} (ln \frac{10}{11} + \frac{2}{11} x - ln - \frac{12}{11} + \frac{2}{11} x))

Simplify

4 (\frac{x ^{2}}{2} + x + 62 (\frac{1}{4} ln 4 x^{2} - 4 x - 120 - \frac{1}{44} (ln \frac{10}{11} + \frac{2}{11} x - ln - \frac{12}{11} + \frac{2}{11} x)) + 30 (- \frac{1}{11} ln ∣ x + 5 ∣ + \frac{1}{11} ln ∣ x - 6 ∣)) - 2 (\frac{1}{4} ln 4 x^{2} - 4 x - 120 - \frac{1}{44} (ln \frac{10}{11} + \frac{2}{11} x - ln - \frac{12}{11} + \frac{2}{11} x)) : 2 x^{2} + 4 x + \frac{123}{2} ln 4 x^{2} - 4 x - 120 - \frac{120}{11} ln ∣ x + 5 ∣ + \frac{120}{11} ln ∣ x - 6 ∣ - \frac{62}{11} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11}) + \frac{1}{22} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11})

= 2 x^{2} + 4 x + \frac{123}{2} ln 4 x^{2} - 4 x - 120 - \frac{120}{11} ln ∣ x + 5 ∣ + \frac{120}{11} ln ∣ x - 6 ∣ - \frac{62}{11} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11}) + \frac{1}{22} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11})

Add a constant to the solution

= 2 x^{2} + 4 x + \frac{123}{2} ln 4 x^{2} - 4 x - 120 - \frac{120}{11} ln ∣ x + 5 ∣ + \frac{120}{11} ln ∣ x - 6 ∣ - \frac{62}{11} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11}) + \frac{1}{22} (ln \frac{2}{11} x + \frac{10}{11} - ln \frac{2}{11} x - \frac{12}{11}) + C

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What is the integral of (4x^3-x)/(x^2-x-30) ?
The integral of (4x^3-x)/(x^2-x-30) is