If 1√5(x-1)^4(x+3)^6dx=A(( x-1)/( x+3))^B+C, where C is the constant of integration, then …

Question

If $\int \frac{1}{\sqrt5(x-1{)}^{4}(x+3{)}^{6}}dx=A{(\frac{\alpha x-1}{\beta x+3})}^{B}+C$, where $C$ is the constant of integration, then the value of $\alpha +\beta +20AB$ is \_\_\_\_\_\_\_\_\_.

Answer: 7

Accepted Answer

$\begin{gathered} I=\int \frac{1}{\sqrt5(x-1{)}^{4}(x+3{)}^{6}}dx \end{gathered}\frac{x+3}{x-1}=t\Rightarrow dx=\frac{-4}{(t-1{)}^{2}}dt\Rightarrow x=(\frac{3+t}{t-1})\Rightarrow (x-1{)}^{4}(x+3{)}^{6}=(x-1{)}^{5}(x+3{)}^{5}(\frac{x+3}{x-1})I=\int \frac{\frac{-4}{(t-1{)}^{2}}dt}{{t}^{1/5}(\frac{3+t}{t-1}-1)(\frac{3+t}{t-3}+3)}I=\int \frac{-4dt}{{t}^{1/5}(16t)}=\frac{5}{4}{(\frac{x-1}{x+3})}^{1/5}+c$

Comparing,

$\begin{gathered} \Rightarrow A=\frac{5}{4},B=\frac{1}{5},\alpha =1,\beta =1 \ \Rightarrow \alpha +\beta +20AB=1+1+20	imes \frac{5}{4}	imes \frac{1}{5}=7 \end{gathered}$

Solutions · 1