What is the integral of (5x^2+7x+6)/(x^2+x+1) ?

The integral of (5x^2+7x+6)/(x^2+x+1) is

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integral of (5x^2+7x+6)/(x^2+x+1)

\int \frac{5 x ^{2} + 7 x + 6}{x ^{2} + x + 1} d x

5 x + ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{7}{3} arctan (\frac{1}{3} (2 x + 1)) + 43 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

\int \frac{5 x ^{2} + 7 x + 6}{x ^{2} + x + 1} d x

Expand

\frac{5 x ^{2} + 7 x + 6}{x ^{2} + x + 1} : \frac{5 x ^{2}}{x ^{2} + x + 1} + \frac{7 x}{x ^{2} + x + 1} + \frac{6}{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{5 x ^{2}}{x ^{2} + x + 1} d x + \int \frac{7 x}{x ^{2} + x + 1} d x + \int \frac{6}{x ^{2} + x + 1} d x

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

\int \frac{7 x}{x ^{2} + x + 1} d x = 14 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3}))

\int \frac{6}{x ^{2} + x + 1} d x = 43 arctan (\frac{1}{3} (2 x + 1))

= 5 (- \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1)) + x) + 14 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + 43 arctan (\frac{1}{3} (2 x + 1))

Simplify

5 (- \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1)) + x) + 14 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + 43 arctan (\frac{1}{3} (2 x + 1)) : 5 x + ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{7}{3} arctan (\frac{1}{3} (2 x + 1)) + 43 arctan (\frac{1}{3} (2 x + 1))

= 5 x + ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{7}{3} arctan (\frac{1}{3} (2 x + 1)) + 43 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= 5 x + ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) - \frac{7}{3} arctan (\frac{1}{3} (2 x + 1)) + 43 arctan (\frac{1}{3} (2 x + 1)) + C

\int (\frac{sec ( x )}{1 + tan ( x )}) d x

\int \frac{x ^{2} + 7 x + 2}{( x + 1 ) ( x - 2 ) ( x + 3 )} d x

\int sin^{2} (\frac{x}{3}) cos^{2} (\frac{x}{3}) d x

\int x^{2} 3 x^{3} + 2 d x

\int \frac{u ^{2}}{u + 1} d u

What is the integral of (5x^2+7x+6)/(x^2+x+1) ?
The integral of (5x^2+7x+6)/(x^2+x+1) is