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__________

Correct answer: 0 aligned & ∫ 1 [5] (x-1)^4(x+3)^6 d x=A((α x-1)/(β x+3))^B+C & I=∫ 1 (x-1)^ 4 / 5 (x+3)^ 6 / 5 d x & I=∫ 1 ((x-1)/(x+3))^ 4 / 5 (x+3)^2 d x & ((x-1)/(x+3))=t (4)/((x+3)^2) d x=d t t^ -4 / 5+1 & I=(1)/(4) ∫ 1 t^ 4 / 5 d t=(1)/(4) t^ 1 / 5 1 / 5 +c & I=(5)/(4)((x-1)/(x+3))^ 1 / 5 +C & A=(5)/(4) α=β=1 B=(1)/(5) & α+β+20 A B=2+20 × (5)/(4) × (1)/(5)=7 aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Indefinite Integration

If ∫ 1 [5] (x-1)^4(x+3)^6 ~d x=A((α x-1)/(β x+3))^B+C , where C is...

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

, where

\mathrm{C}

is the constant of integration, then the value of

\alpha+\beta+20 \mathrm{AB}

is__________

Correct answer

Step-by-step solution

\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}

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