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

by RoRi

Compute the following integral.

    \[ \int \frac{x}{\sqrt{x^2+x+1}} \, dx. \]

First we rearrange the integrand a bit,

    \begin{align*} \int \frac{x}{\sqrt{x^2+x+1}} \, dx &= \frac{1}{2} \int \frac{2x}{\sqrt{x^2+x+1}} \, dx \\[9pt] &= \frac{1}{2} \int \frac{2x+1}{\sqrt{x^2+x+1}} \, dx - \frac{1}{2} \int \frac{1}{\sqrt{x^2+x+1}} \, dx \\[9pt] &= \frac{1}{2} (2 \sqrt{x^2+x+1} ) - \frac{1}{2} \int \frac{1}{\sqrt{\left(x+\frac{1}{2} \right)^2 + \frac{3}{4}}} \, dx. \end{align*}

Now, for the integral on the right, we make the substitution,

    \begin{align*} x + \frac{1}{2} &= \frac{\sqrt{3}}{2} \tan u \\ dx &= \frac{\sqrt{3}}{2} \sec^2 u \, du \\ u &= \arctan \left( \frac{2x+1}{\sqrt{3}} \right). \end{align*}

Therefore, we have

    \begin{align*} \int \frac{x}{\sqrt{x^2+x+1}} \, dx &= \sqrt{x^2+x+1} - \frac{1}{2} \int \frac{1}{\sqrt{\frac{3}{4} \tan^2 u + \frac{3}{4}}} \left( \frac{\sqrt{3}}{2} \sec^2 u \right) \, du \\[9pt] &= \sqrt{x^2+x+1} - \frac{1}{2} \int \frac{sec^2 u \, du}{\sqrt{\tan^2 u + 1}} \\[9pt] &= \sqrt{x^2+x+1} - \frac{1}{2} \int \sec u \, du \\[9pt] &= \sqrt{x^2+x+1} - \frac{1}{2} \log | \tan u + \sec u | + C \\[9pt] &= \sqrt{x^2+x+1} - \frac{1}{2} \log \left| \frac{2x+1}{\sqrt{3}} + \frac{\sqrt{(2x+1)^2 + 3}}{\sqrt{3}} \right| + C \\[9pt] &= \sqrt{x^2+x+1} - \frac{1}{2} \log \left| 2x + 1 + 2 \sqrt{x^2+x+1} \right| + C. \end{align*}

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