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

The integral of (6x^2+3x-2)/((x+1)(x^2+x+1)) is

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

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

ln ∣ x + 1 ∣ + 10 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

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

Take the partial fraction of

\frac{6 x ^{2} + 3 x - 2}{( x + 1 ) ( x ^{2} + x + 1 )} : \frac{1}{x + 1} + \frac{5 x - 3}{x ^{2} + x + 1}

= \int \frac{1}{x + 1} + \frac{5 x - 3}{x ^{2} + x + 1} 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{1}{x + 1} d x + \int \frac{5 x - 3}{x ^{2} + x + 1} d x

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

\int \frac{5 x - 3}{x ^{2} + x + 1} d x = 10 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1))

= ln ∣ x + 1 ∣ + 10 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= ln ∣ x + 1 ∣ + 10 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

\int x^{2} + x + 5 d x

\int \frac{5 x - 8}{x ^{2} - 4} d x

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

\int cos^{2} (2 x + 4) d x

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