What is the integral of (6x^3+4x^2-46x-30)/(x^2-9) ?

The integral of (6x^3+4x^2-46x-30)/(x^2-9) is 3x^2+4x+4ln|x^2-9|-ln| x/3+1|+ln| x/3-1|-27+C

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integral of (6x^3+4x^2-46x-30)/(x^2-9)

\int \frac{6 x ^{3} + 4 x ^{2} - 46 x - 30}{x ^{2} - 9} d x

3 x^{2} + 4 x + 4 ln x^{2} - 9 - ln \frac{x}{3} + 1 + ln \frac{x}{3} - 1 - 27 + C

Solution steps

\int \frac{6 x ^{3} + 4 x ^{2} - 46 x - 30}{x ^{2} - 9} d x

Expand

\frac{6 x ^{3} + 4 x ^{2} - 46 x - 30}{x ^{2} - 9} : \frac{6 x ^{3}}{x ^{2} - 9} + \frac{4 x ^{2}}{x ^{2} - 9} - \frac{46 x}{x ^{2} - 9} - \frac{30}{x ^{2} - 9}

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 ^{3}}{x ^{2} - 9} d x + \int \frac{4 x ^{2}}{x ^{2} - 9} d x - \int \frac{46 x}{x ^{2} - 9} d x - \int \frac{30}{x ^{2} - 9} d x

\int \frac{6 x ^{3}}{x ^{2} - 9} d x = 3 (x^{2} - 9 + 9 ln x^{2} - 9)

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

\int \frac{46 x}{x ^{2} - 9} d x = 23 ln x^{2} - 9

\int \frac{30}{x ^{2} - 9} d x = - 10 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1)

= 3 (x^{2} - 9 + 9 ln x^{2} - 9) - 12 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}) - 23 ln x^{2} - 9 - (- 10 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1))

Simplify

3 (x^{2} - 9 + 9 ln x^{2} - 9) - 12 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}) - 23 ln x^{2} - 9 - (- 10 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1)) : 3 x^{2} + 4 x + 4 ln x^{2} - 9 - ln \frac{x}{3} + 1 + ln \frac{x}{3} - 1 - 27

= 3 x^{2} + 4 x + 4 ln x^{2} - 9 - ln \frac{x}{3} + 1 + ln \frac{x}{3} - 1 - 27

Add a constant to the solution

= 3 x^{2} + 4 x + 4 ln x^{2} - 9 - ln \frac{x}{3} + 1 + ln \frac{x}{3} - 1 - 27 + C

What is the integral of (6x^3+4x^2-46x-30)/(x^2-9) ?
The integral of (6x^3+4x^2-46x-30)/(x^2-9) is 3x^2+4x+4ln|x^2-9|-ln| x/3+1|+ln| x/3-1|-27+C