Let's solve the given integral:

$\int_{1}^{\sqrt{3}} \frac{2}{1 + x^2} \, dx$

Step 1: Recognize the standard integral form

The given integrand, $\frac{1}{1+x^2}$, matches the derivative of the inverse tangent function: $\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}.$

Thus, the integral becomes: $\int \frac{2}{1 + x^2} \, dx = 2 \arctan(x).$

Step 2: Apply limits

Now, evaluate the definite integral: $\int_{1}^{\sqrt{3}} \frac{2}{1 + x^2} \, dx = \left[ 2 \arctan(x) \right]_1^{\sqrt{3}}.$

Substitute the limits into the result: $2 \arctan(\sqrt{3}) - 2 \arctan(1).$

Step 3: Simplify

• $\arctan(\sqrt{3}) = \frac{\pi}{3}$, because $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.
• $\arctan(1) = \frac{\pi}{4}$, because $\tan\left(\frac{\pi}{4}\right) = 1$.

Thus: $2 \arctan(\sqrt{3}) - 2 \arctan(1) = 2 \left(\frac{\pi}{3}\right) - 2 \left(\frac{\pi}{4}\right).$

Step 4: Combine terms

$2 \cdot \frac{\pi}{3} = \frac{2\pi}{3}, \quad 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}.$

Subtract: $\frac{2\pi}{3} - \frac{\pi}{2} = \frac{4\pi}{6} - \frac{3\pi}{6} = \frac{\pi}{6}.$

Final Answer:

$\boxed{\frac{\pi}{6}}$

Would you like further details or have any questions?

Related Questions:

1. How do we determine the antiderivative of trigonometric functions?
2. Can you explain why $\arctan(x)$ is the inverse of the tangent function?
3. How do definite integrals work with limits involving radicals?
4. What are some other standard integral forms to memorize?
5. What is the geometric interpretation of this integral?

Tip:

For integrals involving $\frac{1}{1+x^2}$, always recall the connection to $\arctan(x)$, as it appears frequently in calculus problems.