If ∫ 1 √[5](x-1)^4(x+3)^6 ~d x= A ( α x-1 β x+3 )^B+ C, where C is the constant of integration, then the value of α+β+20 AB is

$$\begin{aligned} & \int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} d x=A\left(\frac{\alpha x-1}{\beta x+3}\right)^B+C \\ & I=\int \frac{1}{(x-1)^{4 / 5}(x+3)^{6 / 5}} d x \\ & I=\int \frac{1}{\left(\frac{x-1}{x+3}\right)^{4 / 5}(x+3)^2} d x \\ & \left(\frac{x-1}{x+3}\right)=t \Rightarrow \frac{4}{(x+3)^2} d x=d t \quad t^{-4 / 5+1} \\ & I=\frac{1}{4} \int \frac{1}{t^{4 / 5}} d t=\frac{1}{4} \frac{t^{1 / 5}}{1 / 5}+c \\ & I=\frac{5}{4}\left(\frac{x-1}{x+3}\right)^{1 / 5}+C \\ & A=\frac{5}{4} \quad \alpha=\beta=1 \quad B=\frac{1}{5} \\ & \alpha+\beta+20 A B=2+20 \times \frac{5}{4} \times \frac{1}{5}=7\end{aligned}$$​∫5(x−1)4(x+3)6​1​dx=A(βx+3αx−1​)B+CI=∫(x−1)4/5(x+3)6/51​dxI=∫(x+3x−1​)4/5(x+3)21​dx(x+3x−1​)=t⇒(x+3)24​dx=dtt−4/5+1I=41​∫t4/51​dt=41​1/5t1/5​+cI=45​(x+3x−1​)1/5+CA=45​α=β=1B=51​α+β+20AB=2+20×45​×51​=7​