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

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

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integral of (x^3-x)/(x^2+4x+9)

\int \frac{x ^{3} - x}{x ^{2} + 4 x + 9} d x

\frac{1}{2} x^{2} - 4 x + \frac{7}{2} ln x^{2} + 4 x + 9 + \frac{24}{5} arctan (\frac{1}{5} (x + 2)) - \frac{1}{2} ln \frac{1}{5} (x + 2)^{2} + 1 - 10 + C

Solution steps

\int \frac{x ^{3} - x}{x ^{2} + 4 x + 9} d x

Expand

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

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 ^{3}}{x ^{2} + 4 x + 9} d x - \int \frac{x}{x ^{2} + 4 x + 9} d x

\int \frac{x ^{3}}{x ^{2} + 4 x + 9} d x = \frac{1}{2} (x + 2)^{2} - 6 (x + 2) + \frac{7}{2} ln x^{2} + 4 x + 9 + \frac{22}{5} arctan (\frac{1}{5} (x + 2))

\int \frac{x}{x ^{2} + 4 x + 9} d x = \frac{1}{2} ln \frac{( x + 2 ) ^{2}}{5} + 1 - \frac{2}{5} arctan (\frac{1}{5} (x + 2))

= \frac{1}{2} (x + 2)^{2} - 6 (x + 2) + \frac{7}{2} ln x^{2} + 4 x + 9 + \frac{22}{5} arctan (\frac{1}{5} (x + 2)) - (\frac{1}{2} ln \frac{( x + 2 ) ^{2}}{5} + 1 - \frac{2}{5} arctan (\frac{1}{5} (x + 2)))

Simplify

\frac{1}{2} (x + 2)^{2} - 6 (x + 2) + \frac{7}{2} ln x^{2} + 4 x + 9 + \frac{22}{5} arctan (\frac{1}{5} (x + 2)) - (\frac{1}{2} ln \frac{( x + 2 ) ^{2}}{5} + 1 - \frac{2}{5} arctan (\frac{1}{5} (x + 2))) : \frac{1}{2} x^{2} - 4 x + \frac{7}{2} ln x^{2} + 4 x + 9 + \frac{24}{5} arctan (\frac{1}{5} (x + 2)) - \frac{1}{2} ln \frac{1}{5} (x + 2)^{2} + 1 - 10

= \frac{1}{2} x^{2} - 4 x + \frac{7}{2} ln x^{2} + 4 x + 9 + \frac{24}{5} arctan (\frac{1}{5} (x + 2)) - \frac{1}{2} ln \frac{1}{5} (x + 2)^{2} + 1 - 10

Add a constant to the solution

= \frac{1}{2} x^{2} - 4 x + \frac{7}{2} ln x^{2} + 4 x + 9 + \frac{24}{5} arctan (\frac{1}{5} (x + 2)) - \frac{1}{2} ln \frac{1}{5} (x + 2)^{2} + 1 - 10 + C

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