What is the integral of (5x^2+2x+6)/(x^3+2x^2+x) ?

The integral of (5x^2+2x+6)/(x^3+2x^2+x) is -ln|x+1|+9/(x+1)+6ln|x|+C

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integral of (5x^2+2x+6)/(x^3+2x^2+x)

\int \frac{5 x ^{2} + 2 x + 6}{x ^{3} + 2 x ^{2} + x} d x

- ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣ + C

Solution steps

\int \frac{5 x ^{2} + 2 x + 6}{x ^{3} + 2 x ^{2} + x} d x

Expand

\frac{5 x ^{2} + 2 x + 6}{x ^{3} + 2 x ^{2} + x} : \frac{5 x ^{2}}{x ^{3} + 2 x ^{2} + x} + \frac{2 x}{x ^{3} + 2 x ^{2} + x} + \frac{6}{x ^{3} + 2 x ^{2} + 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{5 x ^{2}}{x ^{3} + 2 x ^{2} + x} d x + \int \frac{2 x}{x ^{3} + 2 x ^{2} + x} d x + \int \frac{6}{x ^{3} + 2 x ^{2} + x} d x

\int \frac{5 x ^{2}}{x ^{3} + 2 x ^{2} + x} d x = 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1})

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

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

= 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{2}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Simplify

5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{2}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : - ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣

= - ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣ + C

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What is the integral of (5x^2+2x+6)/(x^3+2x^2+x) ?
The integral of (5x^2+2x+6)/(x^3+2x^2+x) is -ln|x+1|+9/(x+1)+6ln|x|+C