Q&A - Ask Doubts and Get Answers
Q

Please,please help me - Integral Calculus - JEE Main-4

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

 
Answers (2)
78 Views

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

Exams
Articles
Questions