If `0 lt x lt 1 ` and `d/(dx){cos^(-1)((1-x)/(1+x))}=1/(f(x){1+f^2(x)})`, then `int(d{f^2(x)})/(f(x)+f^2(x))=`

Options

A. `2ln|1+f^2(x)|+c`
B. `2ln|1+f(x)|+c`
C. `ln|1+f(x)|+c`
D. `ln|1+f^(2) (x)|+c`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \frac{d}{dx} \cos^{-1} \left( \frac{1-x}{1+x} \right) = \frac{1}{f(x)(1+f^2(x))} \]

### Step 2: Simplify the left-hand side
Using the identity \(\cos^{-1}(u) = \theta\) implies \(u = \cos(\theta)\), we can express:
\[ \cos^{-1} \left( \frac{1-x}{1+x} \right) = \theta \implies \frac{1-x}{1+x} = \cos(\theta) \]
This can be rewritten as:
\[ \cos(2\theta) = \frac{1-x}{1+x} \]
Thus, we can differentiate:
\[ \frac{d}{dx} \cos^{-1} \left( \frac{1-x}{1+x} \right) = -\frac{1}{\sqrt{1 - \left( \frac{1-x}{1+x} \right)^2}} \cdot \frac{d}{dx} \left( \frac{1-x}{1+x} \right) \]

### Step 3: Differentiate the right-hand side
Next, we differentiate \(\frac{1-x}{1+x}\):
\[ \frac{d}{dx} \left( \frac{1-x}{1+x} \right) = \frac{-(1+x) - (1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2} \]

### Step 4: Substitute and simplify
Substituting this back, we have:
\[ -\frac{1}{\sqrt{1 - \left( \frac{1-x}{1+x} \right)^2}} \cdot \left( \frac{-2}{(1+x)^2} \right) \]
This simplifies to:
\[ \frac{2}{(1+x)^2 \sqrt{1 - \left( \frac{1-x}{1+x} \right)^2}} \]

### Step 5: Equate the two sides
Now we equate this to the right-hand side:
\[ \frac{2}{(1+x)^2 \sqrt{1 - \left( \frac{1-x}{1+x} \right)^2}} = \frac{1}{f(x)(1+f^2(x))} \]

### Step 6: Find \(f(x)\)
From the above equation, we can deduce that:
\[ f(x) = \sqrt{x} \]
since \(f^2(x) = x\).

### Step 7: Set up the integral
Now we need to evaluate the integral:
\[ \int \frac{d f^2(x)}{f(x) + f^2(x)} \]
Substituting \(f(x) = \sqrt{x}\), we have:
\[ d f^2(x) = d(x) \quad \text{and} \quad f^2(x) = x \]
Thus, the integral becomes:
\[ \int \frac{dx}{\sqrt{x} + x} \]

### Step 8: Simplify the integral
Factoring out \(\sqrt{x}\):
\[ \int \frac{dx}{\sqrt{x}(1 + \sqrt{x})} \]

### Step 9: Use substitution
Let \(t = \sqrt{x}\), then \(x = t^2\) and \(dx = 2t dt\):
\[ \int \frac{2t dt}{t(1+t)} = \int \frac{2 dt}{1+t} \]

### Step 10: Integrate
Integrating gives:
\[ 2 \ln(1+t) + C \]
Substituting back \(t = \sqrt{x}\):
\[ 2 \ln(1+\sqrt{x}) + C \]

### Final Answer
Thus, the final answer is:
\[ 2 \ln(1+\sqrt{x}) + C \]