Given

$F(x)={\displaystyle \underset{1}{\overset{x}{}}}(t^3+t)dt$

Find $F^{\text{'}}(x)$

$F^{\text{'}}(x)=x^3+x$

$F^{\text{'}}(x)=\frac{1}{4}{x}^4+\frac{1}{2}{x}^2$

$F^{\text{'}}(x)=\frac{1}{4}{x}^4+\frac{1}{2}{x}^2-\frac{3}{4}$

$F^{\text{'}}(x)=x^3+x-1$

None of These