Distinct roots of a function involving an integral 
Find the total number of distinct root for $x\in [0,1]$ for which $$\int_0^x\dfrac{t^2}{t^4+1}dt =2x-1$$

The answer is $1$  but I am not able to proceed. 
 A: Sketch:
We want to find the number of zeroes of the function
$$f(x)=\int_0^x\frac{t^2}{t^4+1}\,dt-2x+1$$
in the interval $[0,1]$.
We have
$$f'(x)=\frac{x^2}{x^4+1}-2$$
so $-2\le f'(x)\le -3/2$ for $0\le x\le 1$. This means that $f$ is decreasing, so it can have at most $1$ root.
Finally, note that $f(0)=1$ and
$$f(1)=1+\int_0^1f'(x)\,dx\le1+\int_0^1(-3/2)\,dx=-1/2<0$$
so $f$ changes sign in the interval $[0,1]$, and therefore it has a root there.
A: Define
$$ f(x)=2x-1-\int_0^x\frac{t^2}{t^4+1}\;dt$$
Then $f(0)=-1<0$ and 
$$f(1)=1-\int_0^1\frac{t^2}{t^4+1}\;dt> 0$$
since $\frac{t^2}{t^4+1}<1$ on $[0,1]$.
Therefore $f$ has at least one root on $[0,1]$ by the intermediate value theorem. On the other hand,
$$ f^{\prime}(x)=2-\frac{x^2}{x^4+1}>0$$
on $[0,1]$, so $f$ is increasing and so can have at most one root.
A: Let
$$ F(x) = \int_{0}^{x} \left(2 - \frac{t^2}{t^4 + 1}\right)dt$$
We are looking for roots of $$F(x) = 1$$
Now we have the following inequality
$$ \frac{t^2}{t^4 + 1} \le \frac{1}{2}$$
(follows from $(t^2 -1)^2 \ge 0$)
Thus
$$F'(x) \ge \frac{3}{2}$$
and so $F(x)$ is increasing.
We also have that $$F(1) \ge \int_{0}^{1} \frac{3\ dt}{2} = \frac{3}{2}$$
And $F(0) = 0$
Thus there is exactly one root of $F(x) = 1$ in the interval $[0,1]$
