Integral Calculus Question 289
Question: $ \int_{{}}^{{}}{\log (x+1)dx=} $
[Roorkee 1974]
Options:
A) $ (x+1)\log (x+1)-x+c $
B) $ (x+1)\log (x+1)+x+c $
C) $ (x-1)\log (x+1)-x+c $
D) $ (x-1)\log (x+1)+x+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\log (x+1),dx}=x\log (x+1)-\int_{{}}^{{}}{\frac{x}{x+1},dx+c} $ $ =x\log (x+1)-x+\log (x+1)+c=(x+1)\log (x+1)-x+c $ .
BETA
