JEE Main 2024 — 8 April, Shift 2 — Question 28

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

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

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

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

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

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

$A=\frac{5}{4} \quad \alpha=\beta=1 ,\quad B=\frac{1}{5}$

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