If `f(x)=intsqrt(x)/(x+1)^2dx`, find `f(3)-f(1)`

Answer

To solve the problem, we need to evaluate \( f(3) - f(1) \) where \( f(x) = \int \frac{\sqrt{x}}{(x+1)^2} \, dx \).

### Step 1: Set up the integral
We start with the function:
\[ f(x) = \int \frac{\sqrt{x}}{(x+1)^2} \, dx \]

### Step 2: Use substitution
We will use the substitution \( x = \tan^2 \theta \). Then, we have:
\[ dx = 2 \tan \theta \sec^2 \theta \, d\theta \]
Also, note that:
\[ \sqrt{x} = \tan \theta \]
and
\[ x + 1 = \tan^2 \theta + 1 = \sec^2 \theta \]

### Step 3: Rewrite the integral
Substituting these into the integral gives:
\[ f(x) = \int \frac{\tan \theta}{\sec^4 \theta} \cdot 2 \tan \theta \sec^2 \theta \, d\theta \]
This simplifies to:
\[ f(x) = 2 \int \frac{\tan^2 \theta}{\sec^2 \theta} \, d\theta = 2 \int \sin^2 \theta \, d\theta \]

### Step 4: Integrate \( \sin^2 \theta \)
Using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \):
\[ f(x) = 2 \int \frac{1 - \cos(2\theta)}{2} \, d\theta = \int (1 - \cos(2\theta)) \, d\theta \]
Integrating gives:
\[ f(x) = \theta - \frac{1}{2} \sin(2\theta) + C \]

### Step 5: Back substitute for \( \theta \)
Recall that \( \theta = \tan^{-1}(\sqrt{x}) \). Thus:
\[ f(x) = \tan^{-1}(\sqrt{x}) - \frac{1}{2} \sin(2 \tan^{-1}(\sqrt{x})) + C \]
Using the double angle formula, \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ \sin(2 \tan^{-1}(\sqrt{x})) = \frac{2 \sqrt{x}}{1+x} \]
So,
\[ f(x) = \tan^{-1}(\sqrt{x}) - \frac{\sqrt{x}}{1+x} + C \]

### Step 6: Calculate \( f(3) \) and \( f(1) \)
Now, we need to find \( f(3) \) and \( f(1) \):
1. For \( f(3) \):
\[ f(3) = \tan^{-1}(\sqrt{3}) - \frac{\sqrt{3}}{4} + C = \frac{\pi}{3} - \frac{\sqrt{3}}{4} + C \]
2. For \( f(1) \):
\[ f(1) = \tan^{-1}(1) - \frac{1}{2} + C = \frac{\pi}{4} - \frac{1}{2} + C \]

### Step 7: Compute \( f(3) - f(1) \)
Now, we compute:
\[ f(3) - f(1) = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} + C \right) - \left( \frac{\pi}{4} - \frac{1}{2} + C \right) \]
The constants \( C \) cancel out:
\[ f(3) - f(1) = \frac{\pi}{3} - \frac{\pi}{4} - \frac{\sqrt{3}}{4} + \frac{1}{2} \]
Finding a common denominator (12):
\[ = \frac{4\pi}{12} - \frac{3\pi}{12} - \frac{3\sqrt{3}}{12} + \frac{6}{12} \]
\[ = \frac{\pi + 6 - 3\sqrt{3}}{12} \]

### Final Answer
Thus, the final result is:
\[ f(3) - f(1) = \frac{\pi + 6 - 3\sqrt{3}}{12} \]