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

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

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

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

ln 4 t^{2} - 4 t + 8 + C

Solution steps

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

Expand

\frac{2 t - 1}{t ^{2} - t + 2} : \frac{2 t}{t ^{2} - t + 2} - \frac{1}{t ^{2} - t + 2}

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

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

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

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

Add similar elements:

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

= ln 4 t^{2} - 4 t + 8

Add a constant to the solution

= ln 4 t^{2} - 4 t + 8 + C

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