Integral Calculus Question 303
Question: If $ \int_{{}}^{{}}{\ln (x^{2}+x)dx=x\ln (x^{2}+x)+A} $ , then $ A= $
[MP PET 1992]
Options:
A) $ 2x+\ln (x+1)+ $ constant
B) $ 2x-\ln (x+1)+ $ constant
C) Constant
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\log (x^{2}+x),dx}=\int_{{}}^{{}}{\log x,dx}+\int_{{}}^{{}}{\log (x+1),dx} $ $ =x\log x-x+x\log (x+1)-x+\log (x+1) $ $ =x{ (\log x+\log (x+1) }-2x+\log (x+1) $ $ =x\log (x^{2}+x)-2x+\log (x+1) $ Equating it to the given integration, we get $ A=-2x+\log (x+1) $ .
BETA
