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

The integral of (2x-1)/(x^2-x+3) is ln|4x^2-4x+12|+C

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

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

ln 4 x^{2} - 4 x + 12 + C

Solution steps

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

Expand

\frac{2 x - 1}{x ^{2} - x + 3} : \frac{2 x}{x ^{2} - x + 3} - \frac{1}{x ^{2} - x + 3}

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

\int \frac{2 x}{x ^{2} - x + 3} d x = ln 4 x^{2} - 4 x + 12 + \frac{2}{11} arctan (\frac{2 x - 1}{11})

\int \frac{1}{x ^{2} - x + 3} d x = \frac{2}{11} arctan (\frac{2 x - 1}{11})

= ln 4 x^{2} - 4 x + 12 + \frac{2}{11} arctan (\frac{2 x - 1}{11}) - \frac{2}{11} arctan (\frac{2 x - 1}{11})

Add similar elements:

\frac{2}{11} arctan (\frac{2 x - 1}{11}) - \frac{2}{11} arctan (\frac{2 x - 1}{11}) = 0

= ln 4 x^{2} - 4 x + 12

Add a constant to the solution

= ln 4 x^{2} - 4 x + 12 + C

\int (\frac{7}{x}) d x

\int (x + 5) (x^{2} + 5 x - 10) d x

\frac{\partial}{\partial x} (e^{y} + x + x^{3} + 6)

\frac{d}{d x} (\frac{tan ( x )}{sin ( x )})

u^{2}

What is the integral of (2x-1)/(x^2-x+3) ?
The integral of (2x-1)/(x^2-x+3) is ln|4x^2-4x+12|+C