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The problem asks to evaluate the double integral:

\[
\int_{1}^{\sqrt{2}} \int_{1}^{3} y \cdot \arctan\left(\frac{1}{x}\right) \, dy \, dx
\]

with a hint suggesting the use of **integration by parts**.

### Step-by-Step Solution:

1. **Separate the integrals:**

   The expression is separable because the integrand \( y \cdot \arctan\left(\frac{1}{x}\right) \) can be treated as a product of two independent functions in \( y \) and \( x \). We can rewrite the double integral as:

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

2. **Evaluate the inner integral** over \( y \):

   The inner integral is straightforward as it involves \( y \) alone:

   \[
   \int_{1}^{3} y \, dy = \left[ \frac{y^2}{2} \right]_{1}^{3} = \frac{9}{2} - \frac{1}{2} = 4
   \]

   So, the double integral simplifies to:

   \[
   4 \cdot \int_{1}^{\sqrt{2}} \arctan\left(\frac{1}{x}\right) \, dx
   \]

3. **Integration by parts** for \( \int \arctan\left(\frac{1}{x}\right) \, dx \):

   Use the method of integration by parts. Let:

   \[
   u = \arctan\left(\frac{1}{x}\right), \quad dv = dx
   \]
   Then:

   \[
   du = \frac{-1}{x^2 + 1} \, dx, \quad v = x
   \]

   Applying integration by parts formula:

   \[
   \int u \, dv = uv - \int v \, du
   \]

   This gives:

   \[
   \int \arctan\left(\frac{1}{x}\right) \, dx = x \arctan\left(\frac{1}{x}\right) - \int \frac{x}{x^2 + 1} \, dx
   \]

4. **Simplify the remaining integral:**

   The second integral \( \int \frac{x}{x^2 + 1} \, dx \) is a standard integral:

   \[
   \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \ln(x^2 + 1)
   \]

   Therefore:

   \[
   \int \arctan\left(\frac{1}{x}\right) \, dx = x \arctan\left(\frac{1}{x}\right) - \frac{1}{2} \ln(x^2 + 1)
   \]

5. **Evaluate at the limits \( x = 1 \) and \( x = \sqrt{2} \):**

   Now, substitute the limits into the expression:

   \[
   4 \left[ x \arctan\left(\frac{1}{x}\right) - \frac{1}{2} \ln(x^2 + 1) \right]_{1}^{\sqrt{2}}
   \]

   At \( x = \sqrt{2} \):

   \[
   \sqrt{2} \arctan\left(\frac{1}{\sqrt{2}}\right) - \frac{1}{2} \ln(2 + 1) = \sqrt{2} \cdot \frac{\pi}{6} - \frac{1}{2} \ln(3)
   \]

   At \( x = 1 \):

   \[
   1 \arctan(1) - \frac{1}{2} \ln(1^2 + 1) = \frac{\pi}{4} - \frac{1}{2} \ln(2)
   \]

6. **Final answer:**

   The result of the double integral is:

   \[
   4 \left[ \sqrt{2} \cdot \frac{\pi}{6} - \frac{1}{2} \ln(3) - \left( \frac{\pi}{4} - \frac{1}{2} \ln(2) \right) \right]
   \]

   Simplify the expression to get the exact numerical result.

Would you like further clarification or help with the numerical computation?

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Here are 5 related questions for practice:

1. How do you apply integration by parts to other trigonometric functions?
2. What is the standard form for integrating products of functions?
3. How do you calculate integrals involving inverse trigonometric functions?
4. What is the geometric interpretation of the given integral?
5. Can you find the integral of \( \arctan(x) \) over other limits?

**Tip:** Always check if an integral can be simplified by recognizing separable functions or applying standard integration techniques like substitution or integration by parts.

Find the double integral \( \int_{1}^{\sqrt{2}} \int_{1}^{3} y \cdot \arctan\left(\frac{1}{x}\right) \, dy \, dx \). Hint: Use integration by parts.

Solve a double integral involving the product of a linear term and an inverse trigonometric function. This calculus problem covers the use of integration by parts and standard integrals. Suitable for undergraduate calculus students.

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