Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 322: 80

$-2x+1$

By the Fundamental Theorem of Calculus, the function $f(x)$ is the derivative of $f(x)=3+\int_1^{x^2} \sec (t-1) dt$ is equal to $g'(x)=2x\sec(x^2-1)$ Thus,$g(1)=3; g'(1)=-2$ The linearization of $f(x)$ at $x=-1$ is$L(x)=g(-1)+g'(-1)(x+1)$ Now, $L(x) =3-2(x+1)=-2x+1$