How to solve the integral $\frac{d}{dx} \int_0^{7x^2} \frac{t}{t+66}\ dt$ Can you somebody help me to solve this integral
$$\frac{d}{dx} \int_{0}^{7x^2} \frac{t}{t+66}\, \mathrm{d}t$$
 A: You could integrate, and then differentiate, but that would involve some extra work. And if the thing inside the integral sign was complicated,  probably we could not carry out the integration. 
Instead, use the Fundamental Theorem of Calculus.
Instead of calculating the definite integral, imagine calculating it. You would find a function $F(t)$ whose derivative is $\dfrac{t}{t+66}$, and then you would do the usual substitution: the definite integral is $F(7x^2)-F(0)$.
Now differentiate, using the Chain Rule. We get $14xF'(7x^2)$. But $F'(t)=\dfrac{t}{t+66}$. From this point, I am sure you can finish things.
A: Hint: If $$g(x)=\int_a^{x}f(t)\,dt$$then $$g'(x)=f(x).$$ Now set $h(x)=7x^2$ and notice your equation is $g\circ h(x)$, so apply the chain rule.
A: If $g$ is continuous, and $f(x) = \int_0^x g(t)dt$, then $f'(x) = g(x)$.
In your case, $g(t) = \frac{t}{t+66}$, and we have the function $\phi(x) = f(7 x^2)$. The chain rule gives $\phi'(x) = f'(7 x^2) 14 x = 14 x\,g(7 x^2) = 14x \frac{7 x^2}{7 x^2+66}$.
A: Use DUIS:
Only the first term is active.
\begin{align}
&=\dfrac{7x^2}{7x^2+66}\times 14x
\end{align}
