What is the integral of (12+6x^4)/(x^3+9x) ?

The integral of (12+6x^4)/(x^3+9x) is 3x^2+4/3 ln|x|-83/3 ln|x^2+9|+27+C

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integral of (12+6x^4)/(x^3+9x)

\int \frac{12 + 6 x ^{4}}{x ^{3} + 9 x} d x

3 x^{2} + \frac{4}{3} ln ∣ x ∣ - \frac{83}{3} ln x^{2} + 9 + 27 + C

Solution steps

\int \frac{12 + 6 x ^{4}}{x ^{3} + 9 x} d x

Expand

\frac{12 + 6 x ^{4}}{x ^{3} + 9 x} : \frac{12}{x ^{3} + 9 x} + \frac{6 x ^{4}}{x ^{3} + 9 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{12}{x ^{3} + 9 x} d x + \int \frac{6 x ^{4}}{x ^{3} + 9 x} d x

\int \frac{12}{x ^{3} + 9 x} d x = 12 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9)

\int \frac{6 x ^{4}}{x ^{3} + 9 x} d x = 3 (x^{2} + 9 - 9 ln x^{2} + 9)

= 12 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + 3 (x^{2} + 9 - 9 ln x^{2} + 9)

Simplify

12 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + 3 (x^{2} + 9 - 9 ln x^{2} + 9) : 3 x^{2} + \frac{4}{3} ln ∣ x ∣ - \frac{83}{3} ln x^{2} + 9 + 27

= 3 x^{2} + \frac{4}{3} ln ∣ x ∣ - \frac{83}{3} ln x^{2} + 9 + 27

Add a constant to the solution

= 3 x^{2} + \frac{4}{3} ln ∣ x ∣ - \frac{83}{3} ln x^{2} + 9 + 27 + C

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What is the integral of (12+6x^4)/(x^3+9x) ?
The integral of (12+6x^4)/(x^3+9x) is 3x^2+4/3 ln|x|-83/3 ln|x^2+9|+27+C