What is the integral of (8+4x^4)/(x^3+9x) ?

The integral of (8+4x^4)/(x^3+9x) is 2x^2+8/9 ln|x|-166/9 ln|x^2+9|+18+C

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integral of (8+4x^4)/(x^3+9x)

\int \frac{8 + 4 x ^{4}}{x ^{3} + 9 x} d x

2 x^{2} + \frac{8}{9} ln ∣ x ∣ - \frac{166}{9} ln x^{2} + 9 + 18 + C

Solution steps

\int \frac{8 + 4 x ^{4}}{x ^{3} + 9 x} d x

Expand

\frac{8 + 4 x ^{4}}{x ^{3} + 9 x} : \frac{8}{x ^{3} + 9 x} + \frac{4 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{8}{x ^{3} + 9 x} d x + \int \frac{4 x ^{4}}{x ^{3} + 9 x} d x

\int \frac{8}{x ^{3} + 9 x} d x = 8 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9)

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

= 8 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + 2 (x^{2} + 9 - 9 ln x^{2} + 9)

Simplify

8 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) + 2 (x^{2} + 9 - 9 ln x^{2} + 9) : 2 x^{2} + \frac{8}{9} ln ∣ x ∣ - \frac{166}{9} ln x^{2} + 9 + 18

= 2 x^{2} + \frac{8}{9} ln ∣ x ∣ - \frac{166}{9} ln x^{2} + 9 + 18

Add a constant to the solution

= 2 x^{2} + \frac{8}{9} ln ∣ x ∣ - \frac{166}{9} ln x^{2} + 9 + 18 + C

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What is the integral of (8+4x^4)/(x^3+9x) ?
The integral of (8+4x^4)/(x^3+9x) is 2x^2+8/9 ln|x|-166/9 ln|x^2+9|+18+C