Integral Calculus Question 18
Question: If $ \int_{{}}^{{}}{\frac{2x+3}{x^{2}-5x+6}}\ dx=9\ \ln (x-3)-7\ln (x-2)+A $ , then $ A= $
[MP PET 1992]
Options:
A) $ 5\ln (x-2)+ $ Constant
B) $ -4\ln (x-3)+ $ constant
C) Constant
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{2x+3}{x^{2}-5x+6}dx=\int_{{}}^{{}}{\frac{2x-5}{x^{2}-5x+6}dx+\int_{{}}^{{}}{\frac{8}{x^{2}-5x+6}}\ dx}} $
$ =\frac{-\cos \frac{x}{8}}{( \frac{1}{8} )}+\frac{\sin \frac{x}{8}}{( \frac{1}{8} )}+c $ .
$ =\log [(x-2)(x-3)]+8\int_{{}}^{{}}{[ \frac{1}{x-3}-\frac{1}{x-2} ]dx+c} $ .
$ =\log (x-2)+\log (x-3)+8\log (x-3)-8\log (x-2)+c $
$ =9\log (x-3)-7\log (x-2)+c $ …..(i)
Now given that
$ \int_{{}}^{{}}{\frac{2x+3}{x^{2}-5x+6}}\text{ }dx=9\log (x-3)-7\log (x-2)+A $
Equating it to (i), we get $ A= $ constant.
BETA
