What is the integral of (2x^3-4x^2+18x)/(x^2+9) ?

The integral of (2x^3-4x^2+18x)/(x^2+9) is x^2-4x+12arctan(x/3)+9+C

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integral of (2x^3-4x^2+18x)/(x^2+9)

\int \frac{2 x ^{3} - 4 x ^{2} + 18 x}{x ^{2} + 9} d x

x^{2} - 4 x + 12 arctan (\frac{x}{3}) + 9 + C

Solution steps

\int \frac{2 x ^{3} - 4 x ^{2} + 18 x}{x ^{2} + 9} d x

Expand

\frac{2 x ^{3} - 4 x ^{2} + 18 x}{x ^{2} + 9} : \frac{2 x ^{3}}{x ^{2} + 9} - \frac{4 x ^{2}}{x ^{2} + 9} + \frac{18 x}{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{2 x ^{3}}{x ^{2} + 9} d x - \int \frac{4 x ^{2}}{x ^{2} + 9} d x + \int \frac{18 x}{x ^{2} + 9} d x

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

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

\int \frac{18 x}{x ^{2} + 9} d x = 9 ln x^{2} + 9

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

Simplify

x^{2} + 9 - 9 ln x^{2} + 9 - 12 (- arctan (\frac{x}{3}) + \frac{x}{3}) + 9 ln x^{2} + 9 : x^{2} - 4 x + 12 arctan (\frac{x}{3}) + 9

= x^{2} - 4 x + 12 arctan (\frac{x}{3}) + 9

Add a constant to the solution

= x^{2} - 4 x + 12 arctan (\frac{x}{3}) + 9 + C

What is the integral of (2x^3-4x^2+18x)/(x^2+9) ?
The integral of (2x^3-4x^2+18x)/(x^2+9) is x^2-4x+12arctan(x/3)+9+C