∫5(x−1)4(x+3)61dx=A(βx+3αx−1)B+CI=∫(x−1)4/5(x+3)6/51dxI=∫(x+3x−1)4/5(x+3)21dx(x+3x−1)=t⇒(x+3)24dx=dtt−4/5+1I=41∫t4/51dt=411/5t1/5+cI=45(x+3x−1)1/5+CA=45α=β=1B=51α+β+20AB=2+20×45×51=7
If ∫5(x−1)4(x+3)61 dx=A(βx+3αx−1)B+C, where C is the constant of integration, then the value of α+β+20AB is__________
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