What is the integral of (6x^2+4x+2)/(x^3+x^2+x) ?

The integral of (6x^2+4x+2)/(x^3+x^2+x) is

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integral of (6x^2+4x+2)/(x^3+x^2+x)

\int \frac{6 x ^{2} + 4 x + 2}{x ^{3} + x ^{2} + x} d x

\frac{8}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) + 2 ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1)) + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{6 x ^{2} + 4 x + 2}{x ^{3} + x ^{2} + x} d x

Expand

\frac{6 x ^{2} + 4 x + 2}{x ^{3} + x ^{2} + x} : \frac{6 x ^{2}}{x ^{3} + x ^{2} + x} + \frac{4 x}{x ^{3} + x ^{2} + x} + \frac{2}{x ^{3} + 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{6 x ^{2}}{x ^{3} + x ^{2} + x} d x + \int \frac{4 x}{x ^{3} + x ^{2} + x} d x + \int \frac{2}{x ^{3} + x ^{2} + x} d x

\int \frac{6 x ^{2}}{x ^{3} + x ^{2} + x} d x = 12 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3}))

\int \frac{4 x}{x ^{3} + x ^{2} + x} d x = \frac{8}{3} arctan (\frac{2 x + 1}{3})

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

= 12 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + \frac{8}{3} arctan (\frac{2 x + 1}{3}) + 2 ln ∣ x ∣ - ln 4 x^{2} + 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x + 1))

Simplify

12 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + \frac{8}{3} arctan (\frac{2 x + 1}{3}) + 2 ln ∣ x ∣ - ln 4 x^{2} + 4 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) : \frac{8}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) + 2 ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1)) + 2 ln ∣ x ∣

= \frac{8}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) + 2 ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1)) + 2 ln ∣ x ∣

Add a constant to the solution

= \frac{8}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{2}{3} arctan (\frac{1}{3} (2 x + 1)) + 2 ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1)) + 2 ln ∣ x ∣ + C

\int (cos^{2} (\frac{x}{2})) d x

\int csc^{8} (2 x) d x

\int (x^{3} + x) (x^{4} + 2 x^{2} + 9)^{\frac{1}{2}} d x

\int x (1 - x)^{3} d x

\int x tan^{2} (\frac{x}{2}) d x

What is the integral of (6x^2+4x+2)/(x^3+x^2+x) ?
The integral of (6x^2+4x+2)/(x^3+x^2+x) is