What is the integral of (4x-1)/(2x^2-x+1) ?

The integral of (4x-1)/(2x^2-x+1) is ln|16x^2-8x+8|+C

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integral of (4x-1)/(2x^2-x+1)

\int \frac{4 x - 1}{2 x ^{2} - x + 1} d x

ln 16 x^{2} - 8 x + 8 + C

Solution steps

\int \frac{4 x - 1}{2 x ^{2} - x + 1} d x

Expand

\frac{4 x - 1}{2 x ^{2} - x + 1} : \frac{4 x}{2 x ^{2} - x + 1} - \frac{1}{2 x ^{2} - x + 1}

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

\int \frac{4 x}{2 x ^{2} - x + 1} d x = ln 16 x^{2} - 8 x + 8 + \frac{2}{7} arctan (\frac{4 x - 1}{7})

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

= ln 16 x^{2} - 8 x + 8 + \frac{2}{7} arctan (\frac{4 x - 1}{7}) - \frac{2}{7} arctan (\frac{4 x - 1}{7})

Add similar elements:

\frac{2}{7} arctan (\frac{4 x - 1}{7}) - \frac{2}{7} arctan (\frac{4 x - 1}{7}) = 0

= ln 16 x^{2} - 8 x + 8

Add a constant to the solution

= ln 16 x^{2} - 8 x + 8 + C

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What is the integral of (4x-1)/(2x^2-x+1) ?
The integral of (4x-1)/(2x^2-x+1) is ln|16x^2-8x+8|+C