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Establish the integral formula for (arcsin x)/(x2)

by RoRi

Establish that the following integral formula is correct:

    \[ \int \frac{\arcsin x}{x^2} \, dx = \log \left| \frac{1-\sqrt{1-x^2}}{x} \right| - \frac{\arcsin x}{x} + C. \]

Proof. We can evaluate this integral using integration by parts. Let

    \begin{align*} u &= \arcsin x & du &= \frac{1}{\sqrt{1-x^2}} \, dx \\ dv &= \frac{1}{x^2} \, dx & v &= -\frac{1}{x}. \end{align*}

Then we have

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

Next, we must evaluate the integral on the right. For this integral we have

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

Now, we make a substitution, let u = 1 - \sqrt{1-x^2}, du = \frac{x}{1-x^2} \, dx. Therefore,

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

Combining this with the equation above we then have

    \[ \int \frac{\arcsin x}{x^2} \, dx = \log \left| \frac{1 - \sqrt{1-x^2}}{x} \right| - \frac{\arcsin x}{x} + C. \qquad \blacksquare \]

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