If f((x-4)/(x+2))=2 x+1,(x ∈ R=\ 1,-2\ ) , then ∫ f ( x ) d x is equal to (where C is a constant of integration)

Correct answer: B. -12 _e|1-x|-3 x+c Suppose, (x-4)/(x+2)=y x-4=y(x+2) x(1-y)=2 y+4 x=(2 y+4)/(1-y) So, f(y)=2 ((2 y+4)/(1-y))+1 Now, f(x)=2((2 x+4)/(1-x))+1=(3 x+9)/(1-x) aligned &=(3(x+3))/(1-x)=(3(x-1+4))/(1-x)=-3+(12)/(1-x) & ∫ f(x) d x=-12 _e|1-x|-3 x+c aligned

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Indefinite Integration

If f((x-4)/(x+2))=2 x+1,(x ∈ R=\ 1,-2\ ) , then ∫ f ( x ) d x is...

f\left(\frac{x-4}{x+2}\right)=2 x+1,(x \in R=\{1,-2\})

, then

\int f

(

x

)

d x

is equal to (where

C

is a constant of integration)

Options

A $12 \log _e|1-x|-3 x+c$
B $-12 \log _e|1-x|-3 x+c$
C $-12 \log _e|1-x|+3 x+c$
D $12 \log _e|1-x|+3 x+c$

Correct answer

B. $-12 \log _e|1-x|-3 x+c$

Step-by-step solution

Suppose,

\frac{x-4}{x+2}=y \Rightarrow x-4=y(x+2) \Rightarrow x(1-y)=2 y+4 \Rightarrow x=\frac{2 y+4}{1-y}

So,

f(y)=2

\left(\frac{2 y+4}{1-y}\right)+1

Now,

f(x)=2\left(\frac{2 x+4}{1-x}\right)+1=\frac{3 x+9}{1-x}

\begin{aligned}&=\frac{3(x+3)}{1-x}=\frac{3(x-1+4)}{1-x}=-3+\frac{12}{1-x}\\ &\therefore \int f(x) d x=-12 \log _e|1-x|-3 x+c \end{aligned}

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