What is the integral of (11x+36)/(x^3+6x^2+9x) ?

The integral of (11x+36)/(x^3+6x^2+9x) is 4ln|x|-4ln|x+3|+1/(x+3)+C

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integral of (11x+36)/(x^3+6x^2+9x)

\int \frac{11 x + 36}{x ^{3} + 6 x ^{2} + 9 x} d x

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

Solution steps

\int \frac{11 x + 36}{x ^{3} + 6 x ^{2} + 9 x} d x

Take the partial fraction of

\frac{11 x + 36}{x ^{3} + 6 x ^{2} + 9 x} : \frac{4}{x} - \frac{4}{x + 3} - \frac{1}{( x + 3 ) ^{2}}

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int \frac{4}{x + 3} d x = 4 ln ∣ x + 3 ∣

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (11x+36)/(x^3+6x^2+9x) ?
The integral of (11x+36)/(x^3+6x^2+9x) is 4ln|x|-4ln|x+3|+1/(x+3)+C