Find $F'(x)$ given $ \int_x^{x+2} (4t+1) \ \mathrm{dt}$ Given the problem find $F'(x)$:
$$ \int_x^{x+2} (4t+1) \  \mathrm{dt}$$
I just feel stuck and don't know where to go with this, we learned the second fundamental theorem of calculus today but i don't know where to plug it in. What i did:


*

*chain rule doesn't really take into effect here(*1) so just replace t with $x$

*$F'(x) = 4x + 1$


though the answer is just 8, what am i doing wrong?

Let $g(t)=4t+1$, and let $G(t)$ be an antiderivative of $g(t)$. 
Note that 
$$F(x)=G(x+2)-G(x).\tag{$1$}$$
In this case, we could easily find $G(t)$. But let's not, let's differentiate $F(x)$ immediately. Since $G'(t)=g(t)=4t+1$. we get
$$F'(x)=g(x+2)-g(x)=[4(x+2)+1]-[4x+2].$$ 
This right-hand side ismplifies to $8$. 
