Clear["Global`*"]

expr1 = ((1 - (a*x))^12)*((1 + (a/b))/((1 + (a*x/b))^2)) // Simplify;

Evaluating the integral

int = Assuming[0 < a <= 1 && 0 < b <= 1,
   Integrate[expr1, {x, 0, 1}] // Simplify];

Taking the derivative

expr2 = D[int, a] // Simplify[#, 0 < a <= 1 && 0 < b <= 1] &;

Numerically finding the minimum

NMinimize[{expr2, 0 < a <= 1, 0 < b <= 1}, {a, b}]

(* {-6.00238, {a -> 1.71426*10^-7, b -> 0.0618631}} *)

The minimum occurs when a is near 0

Limit[expr2, a -> 0]

(* -6 *)

Clear["Global`*"]

expr1 = ((1 - (a*x))^12)*((1 + (a/b))/((1 + (a*x/b))^2)) // Simplify;

Evaluating the integral

int = Assuming[0 < a <= 1 && 0 < b <= 1,
   Integrate[expr1, {x, 0, 1}] // Simplify];

Taking the derivative

expr2 = D[int, a] // Simplify[#, 0 < a <= 1 && 0 < b <= 1] &;

Numerically finding the minimum

NMinimize[{expr2, 0 < a <= 1, 0 < b <= 1}, {a, b},
  WorkingPrecision -> 20] // N

{-6., {a -> 8.10042*10^-9, b -> 0.91264}}

Limit[expr2, a -> 0]

(* -6 *)

The minimum occurs when a is near 0

Limit[expr2, a -> 0]

(* -6 *)

EDIT: numerically finding the maximum

NMaximize[{expr2, 0 < a <= 1, 0 < b <= 1}, {a, b},
  WorkingPrecision -> 20] // N

(* {-1.00973*10^-12, {a -> 0.981099, b -> 9.7192*10^-13}} *)

The maximum occurs for b near 0

Limit[expr2, b -> 0]

(* 0 *)

Consequently, expr2 is nonpositive everywhere in the region. Graphically,

Plot3D[expr2, {a, 0, 1}, {b, 0, 1},
 PlotPoints -> 50,
 Exclusions -> True,
 WorkingPrecision -> 15,
 ClippingStyle -> None]