`int_(a^4)^(b^4)(f(sqrtx))/((sqrtx)[f(a^2+b^2-sqrtx)+f(sqrtx)])dx` is equal to

Options

A. `a^(2)-b^(2)`
B. `b^(2)-a^(2)`
C. 0
D. `a^(2)+b^(2)`

Explanation

To solve the integral

\[ I = \int_{a^4}^{b^4} \frac{f(\sqrt{x})}{\sqrt{x} \left( f(a^2 + b^2 - \sqrt{x}) + f(\sqrt{x}) \right)} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( \sqrt{x} = t \). Then, \( x = t^2 \) and \( dx = 2t \, dt \). The limits change as follows:
- When \( x = a^4 \), \( t = a^2 \).
- When \( x = b^4 \), \( t = b^2 \).

Thus, we can rewrite the integral as:

\[ I = \int_{a^2}^{b^2} \frac{f(t)}{t \left( f(a^2 + b^2 - t) + f(t) \right)} \cdot 2t \, dt. \]

### Step 2: Simplifying the Integral
The \( t \) in the numerator and denominator cancels out:

\[ I = 2 \int_{a^2}^{b^2} \frac{f(t)}{f(a^2 + b^2 - t) + f(t)} \, dt. \]

### Step 3: Using the Property of Integrals
We can use the property of integrals which states:

\[ \int_a^b f(x) \, dx = \int_a^b f(b + a - x) \, dx. \]

Applying this property, we have:

\[ I = 2 \int_{a^2}^{b^2} \frac{f(b^2 + a^2 - t)}{f(t) + f(b^2 + a^2 - t)} \, dt. \]

### Step 4: Adding the Two Integrals
Now, we can add the two expressions for \( I \):

\[ 2I = 2 \int_{a^2}^{b^2} \frac{f(t)}{f(a^2 + b^2 - t) + f(t)} \, dt + 2 \int_{a^2}^{b^2} \frac{f(b^2 + a^2 - t)}{f(t) + f(b^2 + a^2 - t)} \, dt. \]

This gives us:

\[ 2I = 2 \int_{a^2}^{b^2} dt = 2(b^2 - a^2). \]

### Step 5: Solving for \( I \)
Dividing both sides by 2, we find:

\[ I = b^2 - a^2. \]

### Conclusion
Thus, the value of the integral is:

\[ \boxed{b^2 - a^2}. \]