What is the integral of (2x-1)/(x^2-x+8) ?

The integral of (2x-1)/(x^2-x+8) is ln|4x^2-4x+32|+C

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integral of (2x-1)/(x^2-x+8)

\int \frac{2 x - 1}{x ^{2} - x + 8} d x

ln 4 x^{2} - 4 x + 32 + C

Solution steps

\int \frac{2 x - 1}{x ^{2} - x + 8} d x

Expand

\frac{2 x - 1}{x ^{2} - x + 8} : \frac{2 x}{x ^{2} - x + 8} - \frac{1}{x ^{2} - x + 8}

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 + 8} d x - \int \frac{1}{x ^{2} - x + 8} d x

\int \frac{2 x}{x ^{2} - x + 8} d x = ln 4 x^{2} - 4 x + 32 + \frac{2}{31} arctan (\frac{2 x - 1}{31})

\int \frac{1}{x ^{2} - x + 8} d x = \frac{2}{31} arctan (\frac{2 x - 1}{31})

= ln 4 x^{2} - 4 x + 32 + \frac{2}{31} arctan (\frac{2 x - 1}{31}) - \frac{2}{31} arctan (\frac{2 x - 1}{31})

Add similar elements:

\frac{2}{31} arctan (\frac{2 x - 1}{31}) - \frac{2}{31} arctan (\frac{2 x - 1}{31}) = 0

= ln 4 x^{2} - 4 x + 32

Add a constant to the solution

= ln 4 x^{2} - 4 x + 32 + C

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What is the integral of (2x-1)/(x^2-x+8) ?
The integral of (2x-1)/(x^2-x+8) is ln|4x^2-4x+32|+C