Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$. $$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\
&= \frac{2x-\sec^2{x}}{\sqrt{2+t^4}} \\
\end{align}$$
Update: 
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$$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+(x^2)^4}}2x - \frac{1}{\sqrt{2+(\tan{x})^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+x^8}} - \frac{\sec^2{x}}{\sqrt{2+\tan^4{x}}} \\
\end{align}$$
Is this correct?


Hint to find derivative: Use FTC, http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements, $F'(x)=\frac{d}{dx}\int^x_a f(t)dt=f(x)$

Answer to original question: HINT to integrate: \begin{align}
\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt &= \frac{1}{\sqrt2} \int_{\tan{x}}^{x^2}  \frac{1}{\sqrt{1+\frac{t^4}{2}}}\,dt \\
&= \frac{1}{\sqrt2} \int_{\tan{x}}^{x^2}  \frac{1}{\sqrt{1+\frac{t^4}{2}}}\,dt \\
\end{align}
Then let $\tan^2 u=t^4/2$
