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If $f\left [ \frac{x-4}{x+2} \right ]=2x+1,\; \left ( x\equiv \mathbf{R}-\left [ 1,-2 \right ] \right )$   then  $\int f\left ( x \right )dx$  is equal to :

( where C is a constant of integration)

• Option 1)

$12\log_{e}\left | 1-x \right |+3x+C$

• Option 2)

$-12\log_{e}\left | 1-x \right |-3x+C$

• Option 3)

$12\log_{e}\left | 1-x \right |-3x+C$

• Option 4)

$-12\log_{e}\left | 1-x \right |+3x+C$

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As we learned

Indefinite integration -

It is inverse process of differentation.

$\frac{d}{dx}\left \{ F(x) \right \}= f(x)$

$\therefore \int f(x)dx= F\left ( x \right )+C$

- wherein

Where

$\frac{d}{dx}F\left ( x \right )$ is differential of $F(x)$ w.r.t  $x$

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

Thus

$\int f\left ( y \right )dy=\frac{2\left ( 2y+4 \right )}{1-y}+1$

$=\int \frac{3\left ( y+1 \right )}{1-y}dy$

$=-\int \frac{12}{1-x}\; \; \; \; -\int 3dx$

$=12\log \left | 1-x \right |-3x+c$

Option 1)

$12\log_{e}\left | 1-x \right |+3x+C$

Option 2)

$-12\log_{e}\left | 1-x \right |-3x+C$

Option 3)

$12\log_{e}\left | 1-x \right |-3x+C$

Option 4)

$-12\log_{e}\left | 1-x \right |+3x+C$

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