Key Concepts and Formulas

• Trigonometric Substitution: Using trigonometric identities to simplify integrals. Specifically, using $x = \tan^2 \theta$ to eliminate the square root.
• Trigonometric Identities: $\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta}$ and $\cos 2\theta = 1 - 2\sin^2 \theta$ (rearranged to $2\sin^2\theta = 1 - \cos 2\theta$).
• Basic Integration: $\int d\theta = \theta + C$ and $\int \cos(a\theta) d\theta = \frac{1}{a} \sin(a\theta) + C$.

Step-by-Step Solution

Step 1: Perform the trigonometric substitution. We are given the integral $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} dx$. Let $x = \tan^2 \theta$, which implies $\sqrt{x} = \tan \theta$ and $dx = 2 \tan \theta \sec^2 \theta d\theta$. This substitution is chosen to simplify the expression under the square root.

Step 2: Substitute and simplify the integral. Substituting into the integral, we get $f(x) = \int \frac{\tan \theta}{(1 + \tan^2 \theta)^2} (2 \tan \theta \sec^2 \theta) d\theta = \int \frac{2 \tan^2 \theta \sec^2 \theta}{(\sec^2 \theta)^2} d\theta = \int \frac{2 \tan^2 \theta}{\sec^2 \theta} d\theta$ Since $\frac{1}{\sec^2 \theta} = \cos^2 \theta$, we have $f(x) = \int 2 \tan^2 \theta \cos^2 \theta d\theta = \int 2 \frac{\sin^2 \theta}{\cos^2 \theta} \cos^2 \theta d\theta = \int 2 \sin^2 \theta d\theta$

Step 3: Use the trigonometric identity to further simplify the integral. Using the identity $2\sin^2 \theta = 1 - \cos 2\theta$, we have $f(x) = \int (1 - \cos 2\theta) d\theta$

Step 4: Integrate with respect to $\theta$. Integrating, we get $f(x) = \int (1 - \cos 2\theta) d\theta = \theta - \frac{\sin 2\theta}{2} + C$

Step 5: Substitute back to express the result in terms of x. Since $x = \tan^2 \theta$, we have $\theta = \arctan(\sqrt{x})$. Also, $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2 \sqrt{x}}{1 + x}$. Therefore, $f(x) = \arctan(\sqrt{x}) - \frac{1}{2} \cdot \frac{2 \sqrt{x}}{1 + x} + C = \arctan(\sqrt{x}) - \frac{\sqrt{x}}{1 + x} + C$

Step 6: Calculate f(3) - f(1). We are asked to find $f(3) - f(1)$. $f(3) = \arctan(\sqrt{3}) - \frac{\sqrt{3}}{1 + 3} + C = \frac{\pi}{3} - \frac{\sqrt{3}}{4} + C$ $f(1) = \arctan(\sqrt{1}) - \frac{\sqrt{1}}{1 + 1} + C = \frac{\pi}{4} - \frac{1}{2} + C$ $f(3) - f(1) = \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4} + C\right) - \left(\frac{\pi}{4} - \frac{1}{2} + C\right) = \frac{\pi}{3} - \frac{\pi}{4} - \frac{\sqrt{3}}{4} + \frac{1}{2} = \frac{4\pi - 3\pi}{12} - \frac{\sqrt{3}}{4} + \frac{1}{2} = \frac{\pi}{12} - \frac{\sqrt{3}}{4} + \frac{1}{2}$ $f(3) - f(1) = \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}$

Common Mistakes & Tips

• Forgetting the constant of integration: Remember to include "+ C" when performing indefinite integration. Although it cancels out when finding the difference f(3) - f(1), it's crucial for the general form of the integral.
• Incorrectly substituting back: Be careful when substituting back from $\theta$ to $x$. Make sure to use the correct trigonometric identities and relationships.
• Sign Errors: Double check your signs when distributing the negative sign during subtraction $f(3) - f(1)$.

Summary

We solved the indefinite integral by using trigonometric substitution, specifically $x = \tan^2 \theta$. We then simplified the integral using trigonometric identities and integrated with respect to $\theta$. Finally, we substituted back to express the result in terms of $x$ and calculated $f(3) - f(1)$, which resulted in $\frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}$.

Final Answer The final answer is $\boxed{\frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}}$, which corresponds to option (B).