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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\frac{d}{d x} (A \cdot e^{x})

f (x) = \frac{x ^{5}}{5} - \frac{4 x ^{3}}{3} + 4 x

f (x) = 7 e^{- 0.03 x}, at x = 9

y = \frac{( 3 x - 9 )}{( 2 x + 8 )}

\int (x^{3} + x) e^{- 4 x} d x

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