JEE Main 2024 (Online) 8th April Evening ShiftQuestion 1
If $\int\frac{1}{\sqrt[5]{(x-1)^{4}(x+3)^{6}}}dx=A\left(\frac{\alpha x-1}{\beta x+3}\right)^{B}+C$, where $C$ is the constant of integration, then the value of $\alpha+\beta+20AB$ is $\_\_\_\_\_\_$ .
Indefinite Integration
Answer$7$
Solution
Concept. The powers $\frac45$ and $\frac65$ add up to $2$; divide by $(x+3)^2$ and use $u=\frac{x-1}{x+3}$. Formulas used. (1) $u=\dfrac{x-1}{x+3}$: $du=\dfrac{4\,dx}{(x+3)^2}$Step 1 - Rewrite. $$\frac{1}{(x-1)^{4/5}(x+3)^{6/5}}=\Big(\frac{x-1}{x+3}\Big)^{-4/5}\frac{1}{(x+3)^2}.$$ Step 2 - Integrate. By (1), $$\int=\frac14\int u^{-4/5}\,du=\frac14\cdot5u^{1/5}+C=\frac54\Big(\frac{x-1}{x+3}\Big)^{1/5}+C.$$ So $A=\frac54$, $B=\frac15$, $\alpha=\beta=1$. Step 3 - Value. $\alpha+\beta+20AB=2+20\cdot\frac14=7$. Answer: 7