Evaluate $\lim_{h\to 0}\frac{1}{h}\int^{x^2+h}_{x^2}\log(3+\arcsin t)\,dt$ I am concerned with how i might evaluate the statement in terms of $x$:
$$\lim_{h\to0}[\frac{1}{h}\int^{x^2+h}_{x^2}\log[3+\arcsin({t})]\,dt$$
On first glance it appears to be just $\log[3+\arcsin({x^2})]$
But is this actually correct? Naturally, if $x$ was a parameter this statement would be true. But what if $x$ is variable?
It is fairly obvious, by first principles of differentiation that 
$\lim_{h\to0}[\frac{1}{h}\int^{(x+h)^2}_{x^2}\log[3+\arcsin({t})]\,dt$] 
is by the fundamental theorem of calculus just $2x\log[3+\arcsin({x^2})$. Is there a similar relationship that can be drawn between the problems?
 A: Consider $$A=\frac{\int^{x^2+h}_{x^2}\log\left(3+\sin^{-1}(t)\right))\,dt }h$$ which is $\frac 00$. Apply L'Hospital with $$u=\int^{x^2+h}_{x^2}\log\left(3+\sin^{-1}(t)\right))\,dt \qquad , \qquad v=h$$ This leads to $$\frac  {u'}{v'}=\frac  {d}{dh}\int^{x^2+h}_{x^2}\log\left(3+\sin^{-1}(t)\right))\,dt $$ Now, use the fundamental theorem of calculus which will give you Jacky Chong's result.
If you have had $$u'=\frac  {d}{dh}\int^{x^2+b(h)}_{x^2+a(h)}\log\left(3+\sin^{-1}(t)\right))\,dt $$ the result would have been $$\frac  {u'}{v'}=b'(h) \log \left(3+\sin ^{-1}\left(x^2+b(h)\right)\right)-a'(h) \log \left(3+\sin
   ^{-1}\left(x^2+a(h)\right)\right)$$ and since $a(h)\to 0$ and $b(h)\to 0$ then $$A \to (b'(h)-a'(h))\log \left(3+\sin ^{-1}\left(x^2\right)\right)$$
Edit
Since you edited the post while I was typing, for the most general case where $$u'=\frac  {d}{dh}\int^{b(x,h)}_{a(x,h)}f(t)\,dt $$ the result would have been $$A \to b'_h(x,h) f(b(x,h))-a'_h(x,h) f(a(x,h))$$ For the second case in the post, then $$u'= 2 (h+x) f\left((h+x)^2\right)$$ making $$A \to 2x\,f(x^2)$$
A: Answer based on your first glance is correct. The problem is solved as a direct application of Fundamental Theorem of Calculus (FTC). Note that the variable of integration is $t$ and the variable of the limit operation is $h$ and hence the expression $x^{2}$ occurring in the integral is a constant and may well be denoted by $a$. So with $a = x^{2}$ we have to find the limit $$L = \lim_{h \to 0}\frac{1}{h}\int_{a}^{a + h}\log(3 + \arcsin t)\,dt\tag{1}$$ Now let's note that if $$F(x) = \int_{a}^{x}f(t)\,dt\tag{2}$$ then via FTC we have $F'(c) = f(c)$ and in particular $F'(a) = f(a)$. Note also by definition of derivative that $$f(a) = F'(a) = \lim_{h \to 0}\frac{F(a + h) - F(a)}{h} = \lim_{h \to 0}\frac{1}{h}\int_{a}^{a + h}f(t)\,dt\tag{3}$$ and now looking at $(1)$ and $(3)$ we can easily see that $$L = f(a) = \log(3 + \arcsin a) = \log(3 + \arcsin x^{2})$$
A: Hint: 
\begin{align}
\frac{1}{h} \int^{x^2+h}_{x^2} \log[3+\arcsin(t)]\ dt \approx  \frac{\log[3+\arcsin(x^2)]}{h} \int^{x^2+h}_{x^2} \ dt=  \log[3+\arcsin(x^2)].
\end{align}
