What is the integral of (2x-1)/(x^2-x+5) ?

The integral of (2x-1)/(x^2-x+5) is ln|4x^2-4x+20|+C

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integral of (2x-1)/(x^2-x+5)

\int \frac{2 x - 1}{x ^{2} - x + 5} d x

ln 4 x^{2} - 4 x + 20 + C

Solution steps

\int \frac{2 x - 1}{x ^{2} - x + 5} d x

Expand

\frac{2 x - 1}{x ^{2} - x + 5} : \frac{2 x}{x ^{2} - x + 5} - \frac{1}{x ^{2} - x + 5}

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 + 5} d x - \int \frac{1}{x ^{2} - x + 5} d x

\int \frac{2 x}{x ^{2} - x + 5} d x = ln 4 x^{2} - 4 x + 20 + \frac{2}{19} arctan (\frac{2 x - 1}{19})

\int \frac{1}{x ^{2} - x + 5} d x = \frac{2}{19} arctan (\frac{2 x - 1}{19})

= ln 4 x^{2} - 4 x + 20 + \frac{2}{19} arctan (\frac{2 x - 1}{19}) - \frac{2}{19} arctan (\frac{2 x - 1}{19})

Add similar elements:

\frac{2}{19} arctan (\frac{2 x - 1}{19}) - \frac{2}{19} arctan (\frac{2 x - 1}{19}) = 0

= ln 4 x^{2} - 4 x + 20

Add a constant to the solution

= ln 4 x^{2} - 4 x + 20 + C

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What is the integral of (2x-1)/(x^2-x+5) ?
The integral of (2x-1)/(x^2-x+5) is ln|4x^2-4x+20|+C