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Evaluate the integral of (3 – x2)1/2 / x

Compute the following integral.

    \[ \int \frac{\sqrt{3-x^2}}{x} \, dx. \]


First, we multiply the numerator and denominator by \sqrt{3-x^2} and do some rearranging to get a friendlier integral,

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

For the integral on the left we make the substitution u = 3-x^2, du = -2x \, dx and obtain

    \[ -\int \frac{x}{\sqrt{3-x^2}} \, dx = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u} = \sqrt{3-x^2}. \]

Then, for the integral on the right we make the substitution u = \sqrt{3-x^2}, du = \frac{-x}{\sqrt{3-x^2}} \, dx. This gives us

    \[ x = \sqrt{3-u^2}, \qquad dx = -\frac{\sqrt{3-x^2}}{x}. \]

Therefore for the integral on the right we have

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

Now, we have to use partial fractions on the integrand,

    \[ \frac{1}{(\sqrt{3} + u)(\sqrt{3} - u)} = \frac{A}{\sqrt{3} + u} + \frac{B}{\sqrt{3} - u}. \]

This gives us the equation

    \[ A(\sqrt{3} - u) + B(\sqrt{3} + u) = 1 \quad \implies \quad A = B = \frac{1}{2\sqrt{3}}. \]

So,

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

Putting these integrals back into our original expression we have

    \begin{align*}  \int \frac{\sqrt{3-x^2}}{x} \, dx &= -\int \frac{x}{\sqrt{3-x^2}} \, dx + 3 \int \frac{1}{x \sqrt{3-x^2}} \, dx \\[10pt]  &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \left( \log | \sqrt{3} + u| - \log | \sqrt{3} - u| \right) \\[10pt]  &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \log \left| \frac{\sqrt{3} + \sqrt{3-x^2}}{\sqrt{3} - \sqrt{3-x^2}} \right| + C \\[10pt]  &= \sqrt{3-x^2} - \frac{\sqrt{3}}{2} \log \left| \frac{(\sqrt{3} + \sqrt{3-x^2})^2}{x^2} \right| + C \\[10pt]  &= \sqrt{3-x^2} - \sqrt{3} \log \left( \frac{\sqrt{3} + \sqrt{3-x^2}}{x} \right) + C. \end{align*}

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