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

Compute the following integral.

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


We have

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

Then we make the substitution

    \begin{align*}  x + \frac{1}{2} &= \frac{1}{2} \sec u \\  dx &= \frac{1}{2} \sec u \tan u \, du \\  u &= \arcsec (2x+1). \end{align*}

Therefore we have

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

(Note: The answer in the book is wrong, it has \log | (2x+1) + 2 \sqrt{x^2+1} |. The \sqrt{x^2+1} should be \sqrt{x^2+x} as we have above.)

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