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

The integral of (x^2)/(x^2+x+2) is -1/2 ln|4x^2+4x+8|-3/(sqrt(7))arctan(1/(sqrt(7))(2x+1))+x+C

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

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

- \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1)) + x + C

Solution steps

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

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

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

\int \frac{- x - 2}{x ^{2} + x + 2} d x = - \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1))

\int 1 d x = x

= - \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1)) + x

Add a constant to the solution

= - \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{3}{7} arctan (\frac{1}{7} (2 x + 1)) + x + C

\int \frac{e ^{x}}{4 - 9 e ^{2 x}} d x

\frac{\partial}{\partial x} (5 - x^{4} + 2 x^{2} - y^{2})

(cos (9 x))^{x}

48 y^{^{''}} + 34 y^{^{'}} - 5 y = 0

\frac{d}{d x} ((4 x - 3)^{2})

What is the integral of (x^2)/(x^2+x+2) ?
The integral of (x^2)/(x^2+x+2) is -1/2 ln|4x^2+4x+8|-3/(sqrt(7))arctan(1/(sqrt(7))(2x+1))+x+C