What is the integral of (x^5-9x^3-9x^2-9)/(x^3+9x) ?

The integral of (x^5-9x^3-9x^2-9)/(x^3+9x) is (x^3)/3-18x-ln|x|-4ln|x^2+9|+54arctan(x/3)+C

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integral of (x^5-9x^3-9x^2-9)/(x^3+9x)

\int \frac{x ^{5} - 9 x ^{3} - 9 x ^{2} - 9}{x ^{3} + 9 x} d x

\frac{x ^{3}}{3} - 18 x - ln ∣ x ∣ - 4 ln x^{2} + 9 + 54 arctan (\frac{x}{3}) + C

Solution steps

\int \frac{x ^{5} - 9 x ^{3} - 9 x ^{2} - 9}{x ^{3} + 9 x} d x

Take the partial fraction of

\frac{x ^{5} - 9 x ^{3} - 9 x ^{2} - 9}{x ^{3} + 9 x} : x^{2} - 18 - \frac{1}{x} + \frac{- 8 x + 162}{x ^{2} + 9}

= \int x^{2} - 18 - \frac{1}{x} + \frac{- 8 x + 162}{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 x^{2} d x - \int 18 d x - \int \frac{1}{x} d x + \int \frac{- 8 x + 162}{x ^{2} + 9} d x

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

\int 18 d x = 18 x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{- 8 x + 162}{x ^{2} + 9} d x = - 4 ln x^{2} + 9 + 54 arctan (\frac{x}{3})

= \frac{x ^{3}}{3} - 18 x - ln ∣ x ∣ - 4 ln x^{2} + 9 + 54 arctan (\frac{x}{3})

Add a constant to the solution

= \frac{x ^{3}}{3} - 18 x - ln ∣ x ∣ - 4 ln x^{2} + 9 + 54 arctan (\frac{x}{3}) + C

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + 1})

\int_{1}^{4} \frac{x ^{2} + 3}{5 x - x ^{2}} d x

n = 1 \sum \infty n n

f (x) = tan (\frac{3 π}{4} x)

\int a^{- x} d x

What is the integral of (x^5-9x^3-9x^2-9)/(x^3+9x) ?
The integral of (x^5-9x^3-9x^2-9)/(x^3+9x) is (x^3)/3-18x-ln|x|-4ln|x^2+9|+54arctan(x/3)+C