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If \(\int \frac{1}{\sqrt[5]{(x-1)^{4}(x+3)^{6}}} d x=A\left(\frac{\alpha x-1}{\beta x+3}\right)^{B}+C\), where \(\mathrm{C}\) is the constant of integration, then the value of \(\alpha+\beta+20 \mathrm{AB}\) is ______.

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Best answer

Correct answer is :

\(\int \frac{1}{\sqrt[5]{(x-1)^{4}(x+3)^{6}}} d x=A\left(\frac{\alpha x-1}{\beta x+3}\right)^{B}+C\)

\(I=\int \frac{1}{(x-1)^{4 / 5}(x+3)^{6 / 5}} d x\)

\(I=\int \frac{1}{\left(\frac{x-1}{x+3}\right)^{4 / 5}(x+3)^{2}} d x\)

\(\left(\frac{\mathrm{x}-1}{\mathrm{x}+3}\right)=\mathrm{t} \Rightarrow \frac{4}{(\mathrm{x}+3)^{2}} \mathrm{dx}=\mathrm{dt} \quad \mathrm{t}^{-4 / 5+1}\)

\(\mathrm{I}=\frac{1}{4} \int \frac{1}{\mathrm{t}^{4 / 5}} \mathrm{dt}=\frac{1}{4} \frac{\mathrm{t}^{1 / 5}}{1 / 5}+\mathrm{c}\)

\(\mathrm{I}=\frac{5}{4}\left(\frac{\mathrm{x}-1}{\mathrm{x}+3}\right)^{1 / 5}+\mathrm{C}\)

\(\mathrm{A}=\frac{5}{4} \quad \alpha=\beta=1 \quad \mathrm{~B}=\frac{1}{5}\)

\(\alpha+\beta+20 \mathrm{AB}=2+20 \times \frac{5}{4} \times \frac{1}{5}=7\) 

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