A) \[y=x\log x-x+2\]
B) \[y=(x+1)\log |x+1|-x+3\]
C) \[y=(x+1)\log |x+1|+x+3\]
D) \[y=x\log x+x+3\]
E) \[y=-(x+1)\log |x+1|+x+3\]
Correct Answer: B
Solution :
\[\frac{dy}{dx}=\log (x+1)\] Þ \[dy=\log (x+1)dx\] \[y=\int{\log (x+1)dx}=x.\log (x+1)-\int{\frac{x}{x+1}dx}\] \[=x.\log (x+1)-\int{\left( 1-\frac{1}{x+1} \right)}\,dx\] \[=x.\log (x+1)-x+\log (x+1)+c=(x+1)\log (x+1)-x+c\] \[x=0\] at \[y=3\] \[3=(1)\log (1)-0+c\] Þ \[3=0+c\] Þ \[c=3\] \ \[y=(x+1)\log |x+1|-x+3\].You need to login to perform this action.
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