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

The integral of (x^2)/(x^2+x+6) is -1/2 ln|4x^2+4x+24|-(11)/(sqrt(23))arctan(1/(sqrt(23))(2x+1))+x+C

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

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

- \frac{1}{2} ln 4 x^{2} + 4 x + 24 - \frac{11}{23} arctan (\frac{1}{23} (2 x + 1)) + x + C

Solution steps

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

\frac{x ^{2}}{x ^{2} + x + 6} = \frac{- x - 6}{x ^{2} + x + 6} + 1

= \int \frac{- x - 6}{x ^{2} + x + 6} + 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{- x - 6}{x ^{2} + x + 6} d x + \int 1 d x

\int \frac{- x - 6}{x ^{2} + x + 6} d x = - \frac{1}{2} ln 4 x^{2} + 4 x + 24 - \frac{11}{23} arctan (\frac{1}{23} (2 x + 1))

\int 1 d x = x

= - \frac{1}{2} ln 4 x^{2} + 4 x + 24 - \frac{11}{23} arctan (\frac{1}{23} (2 x + 1)) + x

Add a constant to the solution

= - \frac{1}{2} ln 4 x^{2} + 4 x + 24 - \frac{11}{23} arctan (\frac{1}{23} (2 x + 1)) + x + C

x + \frac{1}{x ^{3}}

x \to 2 lim (x^{2} - 4)

\frac{d}{d x} (sin^{2} (5 x^{2}))

x \to 1 lim (x - 5)

x \to \infty lim (1 x)

What is the integral of (x^2)/(x^2+x+6) ?
The integral of (x^2)/(x^2+x+6) is -1/2 ln|4x^2+4x+24|-(11)/(sqrt(23))arctan(1/(sqrt(23))(2x+1))+x+C