Answer
$F^{\prime}(x)=8$
Work Step by Step
$F(x)=\displaystyle \int_{x}^{x+2}(4t+1)dt=[4\cdot\frac{1}{2}t^{2}+t]_{x}^{x+2}$
$=[2t^{2}+t]_{x}^{x+2}$
$=[2(x+2)^{2}+(x+2)]-[2x^{2}+x]$
$=2(x^{2}+4x+4)+x+2-2x^{2}-x$
$=2x^{2}+8x+8+x+2-2x^{2}-x$
$F(x)=8x+10$
$F^{\prime}(x)=8$
