Step 1: Understanding the Concept:
The problem provides a function \( f(x) \) defined by an indefinite integral. We are tasked with finding the difference \( f(3) - f(1) \).
According to the Fundamental Theorem of Calculus, if \( f(x) \) is an antiderivative of some function \( g(x) \), then the difference \( f(b) - f(a) \) is equivalent to the definite integral of \( g(x) \) evaluated from the lower limit \( a \) to the upper limit \( b \).
Specifically, \( f(3) - f(1) = \int_{1}^{3} \frac{\sqrt{x}}{(1+x)^2} \, dx \).
This integral is challenging because of the square root in the numerator and the squared term in the denominator.
Key Formula or Approach:
To solve this, we utilize a trigonometric substitution.
The presence of the term \( (1+x) \) suggests a substitution that simplifies using Pythagorean identities, such as \( 1 + \tan^2 \theta = \sec^2 \theta \).
Substituting \( x = \tan^2 \theta \) is ideal because it eliminates the radical \( \sqrt{x} \) and transforms the denominator into a single trigonometric term.
Step 2: Detailed Explanation:
First, let's establish the substitution and the new differential.
Let \( x = \tan^2 \theta \).
Differentiating with respect to \( \theta \), we get:
\[ dx = 2 \tan \theta \sec^2 \theta \, d\theta \]
Next, we must convert the limits of integration from \( x \) to \( \theta \).
For the lower limit, when \( x = 1 \):
\( \tan^2 \theta = 1 \implies \tan \theta = 1 \implies \theta = \frac{\pi}{4} \).
For the upper limit, when \( x = 3 \):
\( \tan^2 \theta = 3 \implies \tan \theta = \sqrt{3} \implies \theta = \frac{\pi}{3} \).
Now substitute these into the integral:
\[ I = \int_{\pi/4}^{\pi/3} \frac{\sqrt{\tan^2 \theta}}{(1 + \tan^2 \theta)^2} \cdot (2 \tan \theta \sec^2 \theta) \, d\theta \]
Simplify the expression inside the integral:
Since \( 1 + \tan^2 \theta = \sec^2 \theta \), we have:
\[ I = \int_{\pi/4}^{\pi/3} \frac{\tan \theta}{(\sec^2 \theta)^2} \cdot (2 \tan \theta \sec^2 \theta) \, d\theta \]
\[ I = \int_{\pi/4}^{\pi/3} \frac{2 \tan^2 \theta \sec^2 \theta}{\sec^4 \theta} \, d\theta = 2 \int_{\pi/4}^{\pi/3} \frac{\tan^2 \theta}{\sec^2 \theta} \, d\theta \]
Convert to sine and cosine terms:
\[ \frac{\tan^2 \theta}{\sec^2 \theta} = \frac{\sin^2 \theta / \cos^2 \theta}{1 / \cos^2 \theta} = \sin^2 \theta \]
The integral becomes:
\[ I = 2 \int_{\pi/4}^{\pi/3} \sin^2 \theta \, d\theta \]
Apply the power-reduction identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \):
\[ I = 2 \int_{\pi/4}^{\pi/3} \frac{1 - \cos 2\theta}{2} \, d\theta = \int_{\pi/4}^{\pi/3} (1 - \cos 2\theta) \, d\theta \]
Performing the integration:
\[ I = \left[ \theta - \frac{\sin 2\theta}{2} \right]_{\pi/4}^{\pi/3} \]
Evaluating at the limits:
Upper limit \( (\theta = \pi/3) \): \( \frac{\pi}{3} - \frac{\sin(2\pi/3)}{2} = \frac{\pi}{3} - \frac{\sqrt{3}/2}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4} \).
Lower limit \( (\theta = \pi/4) \): \( \frac{\pi}{4} - \frac{\sin(\pi/2)}{2} = \frac{\pi}{4} - \frac{1}{2} \).
Subtracting lower from upper:
\[ I = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{3} - \frac{\pi}{4} + \frac{1}{2} - \frac{\sqrt{3}}{4} \]
\[ I = \frac{4\pi - 3\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4} = \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4} \]
Step 3: Final Answer:
The result of the expression is \( \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4} \).