Choose x= -3 as lower limit of integration and pose:
f(x) = 1+ int_(-3)^x (-e^(2t)-2e^t-t)dt
Clearly:
f(-3) = 1+int_(-3)^(-3) (-e^(2t)-2e^t-t)dt = 1
Using now the linearity of integrals:
f(x) = 1 -int_(-3)^x e^(2t)dt -2int_(-3)^xe^tdt - int_(-3)^xtdt
Now:
int_(-3)^x e^(2t)dt = [e^(2t)/2]_(-3)^x = e^(2x)/2-e^-6/2 = e^(2x)-1/(2e^6)
int_(-3)^xe^tdt = [e^t]_(-3)^x = e^x-e^(-3) = e^x-1/e^3
int_(-3)^xtdt = [t^2/2]_(-3)^x = x^2/2 -9/2
Then:
f(x) = 1-e^(2x)/2-2e^x-x^2/2 +1/(2e^6)+2/e^3+9/2
f(x) = (11e^6+4e^3+1)/(2e^6) -(e^(2x)+4e^x+x^2)/2
graph{ (11e^6+4e^3+1)/(2e^6) -(e^(2x)+4e^x+x^2)/2 [-10, 10, -5, 5]}