The total number of distinct $x \in [0, 1]$ for which $\int_{0}^{x} \frac{t^{2}}{1+t^{4}}\,dt = 2x - 1$ is

$\int_0^x \dfrac{t^2}{1+t^4}\,dt = 2x-1, \qquad x \in [0,1]$ $\text{To find: number of distinct solutions } x \in [0,1]$ $1.\ H(0), \ H(1) \text{ signs}$ $2.\ H'(x), \text{ monotonicity}$ $3.\ \text{count the zeros of } H$ $H(x) := \int_0^x \dfrac{t^2}{1+t^4}\,dt - (2x-1)$ $H(0) = 0 - (0-1) = 1 > 0$ $\dfrac{t^2}{1+t^4} \leq \dfrac{1}{2} \text{ on } [0,1] \Rightarrow \int_0^1 \dfrac{t^2}{1+t^4}\,dt \leq \dfrac{1}{2}$ $H(1) = \int_0^1 \dfrac{t^2}{1+t^4}\,dt - 1 \leq -\dfrac{1}{2} < 0$ $H'(x) = \dfrac{x^2}{1+x^4} - 2 \leq \dfrac{1}{2} - 2 = -\dfrac{3}{2} < 0 \Rightarrow H \text{ strictly decreasing}$ $H(0) > 0 > H(1), \ H \text{ strictly decreasing} \Rightarrow \text{exactly one zero in } (0,1)$ $\boxed{\text{Number of solutions} = 1}$