Q.

If ∫1(x-1)4(x+3)65 dx=A(αx-1βx+3)B+C, where C is the constant of integration, then the value of α+β+20AB is ____ .           [2024]


Ans.

(7)

I=∫1(x-1)4(x+3)65 dx

Putting x+3x-1=t ⇒ x=(3+tt-1) ⇒ dx=-4(t-1)2 dt

⇒  (x-1)4(x+3)6=(x-1)5(x+3)5(x+3x-1)

∴   I=∫-4(t-1)2 dtt1/5(3+tt-1-1)(3+tt-1+3)

         I=∫-4 dtt1/5(16t)=-14∫dtt6/5=-14(t-1/5-1/5)+C

On comparing, we get

           A=54,  B=15,  α=1,  β=1

Hence, α+β+20AB=1+1+20×54×15=7