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

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

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integral of (3x^2-x+4)/(x^3+2x^2+2x)

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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