Derivative of definite integral

So, to evaluate $$ \frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt $$ it suffices to note that if $F$ is a primitive (unique up to a constant) of $\frac{t^4 + 1}{t^2 + 1}$, then the integral becomes $F(1) - F(x^2)$. Differentiating in turn gives the result $$ \frac{d}{dx}\int_{x^2}^1\frac{t^4 + 1}{t^2 + 1}dt = \frac{d}{dx} \left( F(1) - F(x^2) \right) = 0 - F'(x^2) \cdot \frac{d}{d x} (x^2) $$ which is simply $$ - 2x \frac{x^8 + 1}{x^4 + 1} . $$

When you calculate the derivative of $$\int_{u(x)}^{v(x)} f(t) dt,$$ what you must realize is that this is actually a compositum function. You have $3$ functions: $F(u,v) = \int_u^v{f(t) dt}$ and the functions $u(x)$ and $v(x)$, giving you. $$\int_{u(x)}^{v(x)} f(t) dt = F(u(x),v(x)).$$ The chain rule for such a function is, \begin{align} \frac{d}{dx}F(u(x),v(x)) &= \frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx} \\&=-f(u(x)) \cdot u'(x) + f(v(x))\cdot v'(x). \end{align}

In your case, $f(t)=\frac{t^4+1}{t^2+1}$, $u(x) = x^2$ and $v(x) = 1$.