What is the integral of (6x^2+23x-9)/(x^3+2x^2-3x) ?

The integral of (6x^2+23x-9)/(x^3+2x^2-3x) is

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integral of (6x^2+23x-9)/(x^3+2x^2-3x)

\int \frac{6 x ^{2} + 23 x - 9}{x ^{3} + 2 x ^{2} - 3 x} d x

6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - \frac{23}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣) + C

Solution steps

\int \frac{6 x ^{2} + 23 x - 9}{x ^{3} + 2 x ^{2} - 3 x} d x

Expand

\frac{6 x ^{2} + 23 x - 9}{x ^{3} + 2 x ^{2} - 3 x} : \frac{6 x ^{2}}{x ^{3} + 2 x ^{2} - 3 x} + \frac{23 x}{x ^{3} + 2 x ^{2} - 3 x} - \frac{9}{x ^{3} + 2 x ^{2} - 3 x}

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{6 x ^{2}}{x ^{3} + 2 x ^{2} - 3 x} d x + \int \frac{23 x}{x ^{3} + 2 x ^{2} - 3 x} d x - \int \frac{9}{x ^{3} + 2 x ^{2} - 3 x} d x

\int \frac{6 x ^{2}}{x ^{3} + 2 x ^{2} - 3 x} d x = 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1))

\int \frac{23 x}{x ^{3} + 2 x ^{2} - 3 x} d x = - \frac{23}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1)

\int \frac{9}{x ^{3} + 2 x ^{2} - 3 x} d x = 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣)

= 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - \frac{23}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣)

Simplify

6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - \frac{23}{4} (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣) : 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - \frac{23}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣)

= 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - \frac{23}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣)

Add a constant to the solution

= 6 (\frac{1}{2} ln x^{2} + 2 x - 3 + \frac{1}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1)) - \frac{23}{4} (ln \frac{1}{2} (x + 1) + 1 - ln \frac{1}{2} (x + 1) - 1) - 9 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣) + C

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What is the integral of (6x^2+23x-9)/(x^3+2x^2-3x) ?
The integral of (6x^2+23x-9)/(x^3+2x^2-3x) is