Integral Calculus Question 159
Question: $ \int_{{}}^{{}}{\frac{x-2}{x^{2}-4x+3}dx=} $
[MP PET 1987]
Options:
A) $ \log \sqrt{x^{2}-4x+3}+c $
B) $ x\log (x-3)-2\log (x-2)+c $
C) $ \log [(x-3)(x-1)] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ x^{2}-4x+3=t\Rightarrow (2x-4),dx=dt $
$ \Rightarrow (x-2),dx=\frac{1}{2}dt, $ then it reduces to $ \frac{1}{2}\int_{{}}^{{}}{\frac{dt}{t}=\frac{1}{2}\log t+c=\frac{1}{2}\log (x^{2}-4x+3)+c.} $
BETA
