A) \[12{{\log }_{e}}|1-x|-3x+c\]
B) \[-12{{\log }_{e}}|1-x|-3x+c\]
C) \[-12{{\log }_{e}}|1-x|+3x+c\]
D) \[12{{\log }_{e}}|1-x|+3x+c\]
Correct Answer: A
Solution :
Let \[\frac{x-4}{x+2}=y\Rightarrow x-4=yx+2y\Rightarrow x(1-y)=2y+4\Rightarrow x=\frac{2y+4}{1-y}\] This gives us\[f(y)=2(\frac{2y+4}{1-y})+1\] So, we have \[f(x)=2(\frac{2x+4}{1-x})+1=\frac{3x+9}{1-x}=-3(\frac{x-1+4}{x-1})=-3-\frac{12}{x-1}\] Thus \[\int_{{}}^{{}}{f(x)dx=12{{\log }_{e}}|1-x|-3x+c}\] So, the correct answer is option A.You need to login to perform this action.
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