What is the integral of (x+4)/(x^3+6x^2+9x) ?

The integral of (x+4)/(x^3+6x^2+9x) is -1/(x+3)+4(1/9 ln|x|-1/9 ln|x+3|+1/(3(x+3)))+C

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integral of (x+4)/(x^3+6x^2+9x)

\int \frac{x + 4}{x ^{3} + 6 x ^{2} + 9 x} d x

- \frac{1}{x + 3} + 4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{3 ( x + 3 )}) + C

Solution steps

\int \frac{x + 4}{x ^{3} + 6 x ^{2} + 9 x} d x

Expand

\frac{x + 4}{x ^{3} + 6 x ^{2} + 9 x} : \frac{x}{x ^{3} + 6 x ^{2} + 9 x} + \frac{4}{x ^{3} + 6 x ^{2} + 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{x}{x ^{3} + 6 x ^{2} + 9 x} d x + \int \frac{4}{x ^{3} + 6 x ^{2} + 9 x} d x

\int \frac{x}{x ^{3} + 6 x ^{2} + 9 x} d x = - \frac{1}{x + 3}

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

= - \frac{1}{x + 3} + 4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{3 ( x + 3 )})

Add a constant to the solution

= - \frac{1}{x + 3} + 4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{3 ( x + 3 )}) + C

What is the integral of (x+4)/(x^3+6x^2+9x) ?
The integral of (x+4)/(x^3+6x^2+9x) is -1/(x+3)+4(1/9 ln|x|-1/9 ln|x+3|+1/(3(x+3)))+C