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

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

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

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

3 ln ∣ x + 1 ∣ + 3 ln x^{2} + 1 + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= 3 ln ∣ x + 1 ∣ + 3 ln x^{2} + 1

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ + 3 ln x^{2} + 1 + C

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