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The given integral is:

\[
\int_{1}^{\sqrt{3}} \arctan\left(\frac{1}{x}\right) \frac{1}{x} \, dx
\]

To compute this definite integral, let's proceed with the following steps:

### Step 1: Substitution
We'll start by substituting \( u = \frac{1}{x} \). Thus, \( du = -\frac{1}{x^2} dx \) or equivalently \( dx = -\frac{1}{u^2} du \).

When \( x = 1 \), \( u = 1 \).

When \( x = \sqrt{3} \), \( u = \frac{1}{\sqrt{3}} \).

Substituting into the integral, we get:

\[
\int_{1}^{\sqrt{3}} \arctan\left(\frac{1}{x}\right) \frac{1}{x} \, dx = \int_{1}^{\frac{1}{\sqrt{3}}} \arctan(u) \cdot (-u^2) \, du
\]

### Step 2: Simplifying the Integral
The integral becomes:

\[
-\int_{1}^{\frac{1}{\sqrt{3}}} u \arctan(u) \, du
\]

Since the limits of integration are reversed, we can switch the bounds and remove the negative sign:

\[
\int_{\frac{1}{\sqrt{3}}}^{1} u \arctan(u) \, du
\]

### Step 3: Integration by Parts
Now, apply integration by parts. Let:
- \( v = \arctan(u) \) so that \( dv = \frac{1}{1+u^2} du \)
- \( dw = u du \) so that \( w = \frac{u^2}{2} \)

The integration by parts formula is:

\[
\int v \, dw = v \cdot w - \int w \, dv
\]

Substituting into the formula:

\[
\int_{\frac{1}{\sqrt{3}}}^{1} u \arctan(u) \, du = \left. \frac{u^2}{2} \arctan(u) \right|_{\frac{1}{\sqrt{3}}}^{1} - \int_{\frac{1}{\sqrt{3}}}^{1} \frac{u^2}{2} \cdot \frac{1}{1+u^2} \, du
\]

### Step 4: Simplifying the Second Integral
Notice that:

\[
\frac{u^2}{1+u^2} = 1 - \frac{1}{1+u^2}
\]

So the second integral simplifies to:

\[
\frac{1}{2} \int_{\frac{1}{\sqrt{3}}}^{1} \left( 1 - \frac{1}{1+u^2} \right) du
\]

This can be split into two integrals:

\[
\frac{1}{2} \left( \int_{\frac{1}{\sqrt{3}}}^{1} 1 \, du - \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{1+u^2} \, du \right)
\]

The first integral is straightforward:

\[
\frac{1}{2} \left( u \Big|_{\frac{1}{\sqrt{3}}}^{1} \right) = \frac{1}{2} \left( 1 - \frac{1}{\sqrt{3}} \right)
\]

The second integral is the arctangent function again:

\[
\frac{1}{2} \left( \arctan(u) \Big|_{\frac{1}{\sqrt{3}}}^{1} \right) = \frac{1}{2} \left( \arctan(1) - \arctan\left(\frac{1}{\sqrt{3}}\right) \right)
\]

### Final Answer
Putting everything together, the definite integral evaluates to:

\[
\frac{1}{2} \left( \frac{1}{2} - \frac{1}{2\sqrt{3}} - \arctan\left(\frac{1}{\sqrt{3}}\right) + \frac{\pi}{4} \right) + \frac{1}{2} \left(\arctan(1) - \arctan\left(\frac{1}{\sqrt{3}}\right)\right)
\]

Simplify further to get the final numerical result:

\[
\frac{1}{4} \left(2\arctan(1) - \arctan\left(\frac{1}{\sqrt{3}}\right)\right) + \frac{1}{4} - \frac{1}{4\sqrt{3}}
\]

This is the exact value of the integral.

Do you need any further details or have any questions?  
Here are 8 related questions you might find interesting:

1. How do you perform integration by parts in general?
2. What are the properties of the arctan function?
3. How does substitution simplify integrals?
4. Why is the integral of \(\frac{1}{1+u^2}\) equal to arctan(u)?
5. How can definite integrals be interpreted geometrically?
6. What are some common strategies for solving definite integrals?
7. How do you handle integrals with inverse trigonometric functions?
8. What other methods can be used to evaluate difficult integrals?

**Tip:** When evaluating definite integrals involving inverse trigonometric functions, consider using trigonometric identities or substitutions to simplify the expression.

Learn how to compute the definite integral ∫(arctan(1/x) / x) dx from 1 to √3 step-by-step. This problem involves integration techniques including substitution and integration by parts. Suitable for college-level mathematics.

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