What is the integral of (x^4)/(3x^2+27) ?

The integral of (x^4)/(3x^2+27) is (x^3)/9-3x+9arctan(x/3)+C

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integral of (x^4)/(3x^2+27)

\int \frac{x ^{4}}{3 x ^{2} + 27} d x

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

Solution steps

\int \frac{x ^{4}}{3 x ^{2} + 27} d x

Long division

\frac{x ^{4}}{3 x ^{2} + 27} : \frac{x ^{2}}{3} - 3 + \frac{27}{x ^{2} + 9}

= \int \frac{x ^{2}}{3} - 3 + \frac{27}{x ^{2} + 9} d 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{x ^{2}}{3} d x - \int 3 d x + \int \frac{27}{x ^{2} + 9} d x

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

\int 3 d x = 3 x

\int \frac{27}{x ^{2} + 9} d x = 9 arctan (\frac{x}{3})

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

Add a constant to the solution

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

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