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I have seen that the integral $\int x \sec^{-1}(x) dx$ is evaluated in the following way using partial integration:

$$\int x \sec^{-1}(x) dx =\dfrac{x^2}{2}\sec^{-1}(x)-{\displaystyle\int}\dfrac{x}{2\sqrt{x^2-1}}\,\mathrm{d}x$$

That is, $\frac{d}{dx} \sec^{-1}(x)=\frac{1}{x\sqrt{x^2-1}}$ is used.

However, it is actually known that $\frac{d}{dx} \sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}$.

Why can the absolute value signs be omitted?

EDIT: I have also seen this integral but it doesn't solve my problem:

$$\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-1}}}}}}\,=\text{arcsec}\,\left| u \right|+C$$

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1 Answer 1

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As you have suspected, the absolute value cannot be omitted. So the proposed answer only works for $x > 1$.

For a more precise computation, we may recognize the derivative of $\operatorname{arcsec}$ as

$$ \frac{d}{dz}\operatorname{arcsec}(z) = \frac{1}{z^2\sqrt{1-\frac{1}{z^2}}}. $$

This formula holds on $\{z\in\mathbb{C} : |z| > 1 \}$. Then we can compute the integral as

\begin{align} \int z \operatorname{arcsec}(z) \, \mathrm{d}z &= \frac{z^2}{2}\operatorname{arcsec}(z) - \int \frac{1}{2\sqrt{1-\frac{1}{z^2}}} \, \mathrm{d}z \\ &= \frac{z^2}{2}\operatorname{arcsec}(z) - \frac{z}{2}\sqrt{1 - \frac{1}{z^2}} + \mathsf{C}. \end{align}

Again, this holds on all of $\{z \in \mathbb{C} : |z| > 1\}$, and in particular, this reduces to

$$ \int x \operatorname{arcsec}(x) \, \mathrm{d}x = \frac{x^2}{2}\operatorname{arcsec}(x) - \frac{\operatorname{sign}(x)}{2}\sqrt{x^2 - 1} + \mathsf{C} $$

for real $x$. Obviously this is different from what we get when we compute the first line in OP's question.

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