The total number for distinct `x epsilon[0,1]` for which `int_(0)^(x)(t^(2))/(1+t^(4))dt=2x-1` is __________.

Explanation

To solve the equation

\[ \int_{0}^{x} \frac{t^2}{1 + t^4} dt = 2x - 1 \]

for distinct values of \( x \) in the interval \([0, 1]\), we will define a function \( f(x) \) as follows:

\[ f(x) = \int_{0}^{x} \frac{t^2}{1 + t^4} dt + 1 - 2x \]

We need to find the values of \( x \) for which \( f(x) = 0 \).

### Step 1: Evaluate \( f(0) \)

Calculating \( f(0) \):

\[ f(0) = \int_{0}^{0} \frac{t^2}{1 + t^4} dt + 1 - 2(0) = 0 + 1 = 1 \]

### Step 2: Find the derivative \( f'(x) \)

To find \( f'(x) \), we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \int_{0}^{x} \frac{t^2}{1 + t^4} dt \right) - 2 \]

Using the Fundamental Theorem of Calculus, we have:

\[ f'(x) = \frac{x^2}{1 + x^4} - 2 \]

### Step 3: Analyze \( f'(x) \)

We need to analyze the sign of \( f'(x) \):

- The term \( \frac{x^2}{1 + x^4} \) is always non-negative for \( x \in [0, 1] \).
- The maximum value of \( \frac{x^2}{1 + x^4} \) occurs at \( x = 1 \):

\[ f'(1) = \frac{1^2}{1 + 1^4} - 2 = \frac{1}{2} - 2 = -\frac{3}{2} \]

Since \( f'(x) \) is negative at \( x = 1 \) and \( f'(x) \) is a continuous function, we can conclude that \( f'(x) < 0 \) for all \( x \in [0, 1] \). Thus, \( f(x) \) is a decreasing function.

### Step 4: Evaluate \( f(1) \)

Now, let's find \( f(1) \):

\[ f(1) = \int_{0}^{1} \frac{t^2}{1 + t^4} dt + 1 - 2(1) \]

We need to evaluate the integral:

\[ \int_{0}^{1} \frac{t^2}{1 + t^4} dt \]

This integral is positive and less than 1, as we can see from the following:

- The integrand \( \frac{t^2}{1 + t^4} \) is always less than 1 for \( t \in [0, 1] \).

Thus, we can conclude:

\[ f(1) < 0 \]

### Step 5: Conclusion about the number of solutions

Since \( f(0) = 1 > 0 \) and \( f(1) < 0 \), and knowing that \( f(x) \) is a decreasing function, by the Intermediate Value Theorem, there must be exactly one value of \( x \) in the interval \([0, 1]\) where \( f(x) = 0 \).

Therefore, the total number of distinct \( x \) in the interval \([0, 1]\) that satisfies the equation is:

\[ \boxed{1} \]