The total number for distinct `x in[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 based on the left-hand side and analyze its behavior.

### Step 1: Define the function
Let

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

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

### Step 2: Analyze the function \( f(x) \)
To understand the behavior of \( f(x) \), we will compute its derivative \( f'(x) \):

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

Using the Fundamental Theorem of Calculus, we have:

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

### Step 3: Determine the sign of \( f'(x) \)
Now, we need to analyze the sign of \( f'(x) \):

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

Setting \( f'(x) = 0 \) gives:

\[ \frac{x^2}{1+x^4} = 2 \implies x^2 = 2(1+x^4) \implies 2x^4 - x^2 + 2 = 0 \]

Let \( y = x^2 \). Then we have:

\[ 2y^2 - y + 2 = 0 \]

### Step 4: Analyze the quadratic equation
The discriminant of the quadratic equation \( 2y^2 - y + 2 = 0 \) is:

\[ D = (-1)^2 - 4 \cdot 2 \cdot 2 = 1 - 16 = -15 \]

Since the discriminant is negative, there are no real roots for \( f'(x) = 0 \). This means \( f'(x) \) does not change sign.

### Step 5: Evaluate \( f'(x) \) at the endpoints
Now we check the sign of \( f'(x) \) at the endpoints of the interval \([0, 1]\):

- At \( x = 0 \):

\[ f'(0) = \frac{0^2}{1+0^4} - 2 = -2 < 0 \]

- At \( x = 1 \):

\[ f'(1) = \frac{1^2}{1+1^4} - 2 = \frac{1}{2} - 2 = -\frac{3}{2} < 0 \]

Since \( f'(x) < 0 \) for all \( x \in [0, 1] \), it follows that \( f(x) \) is a decreasing function.

### Step 6: Evaluate \( f(0) \) and \( f(1) \)
Now we evaluate \( f(0) \) and \( f(1) \):

- At \( x = 0 \):

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

- At \( x = 1 \):

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

Since \( \frac{t^2}{1+t^4} \) is positive for \( t \in [0, 1] \), the integral \( \int_{0}^{1} \frac{t^2}{1+t^4} dt \) is positive. Therefore, \( f(1) < 0 \).

### Step 7: Conclusion
Since \( f(0) = 1 > 0 \) and \( f(1) < 0 \), and \( f(x) \) is continuous and decreasing, by the Intermediate Value Theorem, there is exactly one solution \( x \) in the interval \([0, 1]\) where \( f(x) = 0 \).

Thus, the total number of distinct \( x \) in \([0, 1]\) for which

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

is

\[ \boxed{1} \]