What is the integral of 1/(x^3+6x^2+9x) ?

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

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integral of 1/(x^3+6x^2+9x)

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{3} + 6 x ^{2} + 9 x} : \frac{1}{9 x} - \frac{1}{9 ( x + 3 )} - \frac{1}{3 ( x + 3 ) ^{2}}

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

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{1}{9 ( x + 3 )} d x = \frac{1}{9} ln ∣ x + 3 ∣

\int \frac{1}{3 ( x + 3 ) ^{2}} d x = - \frac{1}{3 ( x + 3 )}

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

Simplify

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

Add a constant to the solution

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

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