What is the integral of ((x-9))/(x^3+4x^2+16x) ?

The integral of ((x-9))/(x^3+4x^2+16x) is

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integral of ((x-9))/(x^3+4x^2+16x)

\int \frac{( x - 9 )}{x ^{3} + 4 x ^{2} + 16 x} d x

\frac{1}{2 3} arctan (\frac{x + 2}{2 3}) - 9 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3})) + C

Solution steps

\int \frac{x - 9}{x ^{3} + 4 x ^{2} + 16 x} d x

Expand

\frac{x - 9}{x ^{3} + 4 x ^{2} + 16 x} : \frac{x}{x ^{3} + 4 x ^{2} + 16 x} - \frac{9}{x ^{3} + 4 x ^{2} + 16 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}{x ^{3} + 4 x ^{2} + 16 x} d x - \int \frac{9}{x ^{3} + 4 x ^{2} + 16 x} d x

\int \frac{x}{x ^{3} + 4 x ^{2} + 16 x} d x = \frac{1}{2 3} arctan (\frac{x + 2}{2 3})

\int \frac{9}{x ^{3} + 4 x ^{2} + 16 x} d x = 9 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3}))

= \frac{1}{2 3} arctan (\frac{x + 2}{2 3}) - 9 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3}))

Add a constant to the solution

= \frac{1}{2 3} arctan (\frac{x + 2}{2 3}) - 9 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{32} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3})) + C

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What is the integral of ((x-9))/(x^3+4x^2+16x) ?
The integral of ((x-9))/(x^3+4x^2+16x) is