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) \[(x-1)\log (x+1)+x+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\log (x+1)\,dx}=x\log (x+1)-\int_{{}}^{{}}{\frac{x}{x+1}\,dx+c}\] \[=x\log (x+1)-x+\log (x+1)+c=(x+1)\log (x+1)-x+c\].
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