If 2 x^2+3 (x^2-1 ) (x^2-4 ) d x= [ ( x-2 x+2 )^a ( x+1 x-1 )^b ]+c , (where c is the constant of integration) then the value of a+ b is equal to

Step-by-step solution

Let I = 2x^2+3 (x^2-1)(x^2-4) dx . Using the substitution y = x^2 , we rewrite the integrand as 2y+3 (y-1)(y-4) and decompose it into partial fractions: 2y+3 (y-1)(y-4) = A y-1 + B y-4 Multiplying both sides by (y-1)(y-4) gives 2y+3 = A(y-4) + B(y-1) . Substituting y=1 yields 5 = -3A , so A = - 5 3 . Substituting y=4 gives 11 = 3B , so B = 11 3 . The integral becomes: I = - 5 3 1 x^2-1 dx + 11 3 1 x^2-4 dx Using the standard integral 1 x^2-a^2 dx = 1 2a | x-a x+a | + C : - 5 3 1 2 | x-1 x+1 | = 5 6 | x+1 x-1 | 11 3