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

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

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

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

\frac{3}{2} x^{2} + \frac{2}{3} ln ∣ x ∣ - \frac{83}{6} ln x^{2} + 9 + \frac{27}{2} + C

Solution steps

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

Expand

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

\int \frac{6}{x ^{3} + 9 x} d x = 6 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9)

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

= 6 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + \frac{3}{2} (x^{2} + 9 - 9 ln x^{2} + 9)

Simplify

6 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + \frac{3}{2} (x^{2} + 9 - 9 ln x^{2} + 9) : \frac{3}{2} x^{2} + \frac{2}{3} ln ∣ x ∣ - \frac{83}{6} ln x^{2} + 9 + \frac{27}{2}

= \frac{3}{2} x^{2} + \frac{2}{3} ln ∣ x ∣ - \frac{83}{6} ln x^{2} + 9 + \frac{27}{2}

Add a constant to the solution

= \frac{3}{2} x^{2} + \frac{2}{3} ln ∣ x ∣ - \frac{83}{6} ln x^{2} + 9 + \frac{27}{2} + C

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