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Evaluate the integral of 1 / (x2 – x + 2)

by RoRi

Evaluate the following integral

    \[ \int \frac{dx}{x^2-x+2}. \]

First, we want to rewrite the term in the denominator,

    \[ x^2 - x + 2 = \frac{7}{4} + \left(x - \frac{1}{2} \right)^2. \]

Then we have

    \begin{align*} \int \frac{dx}{x^2-x+2} &= \int \frac{dx}{\frac{7}{4} + \left( x - \frac{1}{2} \right)^2} \\[9pt] &= \frac{4}{7} \int \frac{dx}{1+ \left( \frac{2x-1}{\sqrt{7}} \right)^2}. \end{align*}

Now, let u = \frac{2x-1}{\sqrt{7}} which implies du = \frac{2}{\sqrt{7}} \, dx. Therefore,

    \begin{align*} \int \frac{dx}{x^2-x+2} &= \frac{4}{7} \int \frac{dx}{1+\left( \frac{2x-1}{\sqrt{7}} \right)^2} \\ &= \frac{2}{\sqrt{7}} \int \frac{du}{1+u^2} \\ &= \frac{2}{\sqrt{7}} \arctan u + C \\ &= \frac{2}{\sqrt{7}} \arctan \left( \frac{2x-1}{\sqrt{7}} \right) + C. \end{align*}

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