What is the integral of (2x^4+6x^2-108)/(x^2+3x) ?

The integral of (2x^4+6x^2-108)/(x^2+3x) is (2x^3)/3-3x^2+24x-36ln|x+3|-36ln|x|+18+C

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integral of (2x^4+6x^2-108)/(x^2+3x)

\int \frac{2 x ^{4} + 6 x ^{2} - 108}{x ^{2} + 3 x} d x

\frac{2 x ^{3}}{3} - 3 x^{2} + 24 x - 36 ln ∣ x + 3 ∣ - 36 ln ∣ x ∣ + 18 + C

Solution steps

\int \frac{2 x ^{4} + 6 x ^{2} - 108}{x ^{2} + 3 x} d x

Expand

\frac{2 x ^{4} + 6 x ^{2} - 108}{x ^{2} + 3 x} : \frac{2 x ^{4}}{x ^{2} + 3 x} + \frac{6 x ^{2}}{x ^{2} + 3 x} - \frac{108}{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{2 x ^{4}}{x ^{2} + 3 x} d x + \int \frac{6 x ^{2}}{x ^{2} + 3 x} d x - \int \frac{108}{x ^{2} + 3 x} d x

\int \frac{2 x ^{4}}{x ^{2} + 3 x} d x = \frac{2 x ^{3}}{3} - 3 x^{2} + 18 x - 54 ln ∣ x + 3 ∣

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

\int \frac{108}{x ^{2} + 3 x} d x = 108 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣)

= \frac{2 x ^{3}}{3} - 3 x^{2} + 18 x - 54 ln ∣ x + 3 ∣ + 6 (x + 3 - 3 ln ∣ x + 3 ∣) - 108 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣)

Simplify

\frac{2 x ^{3}}{3} - 3 x^{2} + 18 x - 54 ln ∣ x + 3 ∣ + 6 (x + 3 - 3 ln ∣ x + 3 ∣) - 108 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣) : \frac{2 x ^{3}}{3} - 3 x^{2} + 24 x - 36 ln ∣ x + 3 ∣ - 36 ln ∣ x ∣ + 18

= \frac{2 x ^{3}}{3} - 3 x^{2} + 24 x - 36 ln ∣ x + 3 ∣ - 36 ln ∣ x ∣ + 18

Add a constant to the solution

= \frac{2 x ^{3}}{3} - 3 x^{2} + 24 x - 36 ln ∣ x + 3 ∣ - 36 ln ∣ x ∣ + 18 + C

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What is the integral of (2x^4+6x^2-108)/(x^2+3x) ?
The integral of (2x^4+6x^2-108)/(x^2+3x) is (2x^3)/3-3x^2+24x-36ln|x+3|-36ln|x|+18+C