A) \[2x+\ln (x+1)+\]constant
B) \[2x-\ln (x+1)+\]constant
C) Constant
D) None of these
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\log ({{x}^{2}}+x)\,dx}=\int_{{}}^{{}}{\log x\,dx}+\int_{{}}^{{}}{\log (x+1)\,dx}\] \[=x\log x-x+x\log (x+1)-x+\log (x+1)\] \[=x\left\{ (\log x+\log (x+1) \right\}-2x+\log (x+1)\] \[=x\log ({{x}^{2}}+x)-2x+\log (x+1)\] Equating it to the given integration, we get \[A=-2x+\log (x+1)\].
You need to login to perform this action.
You will be redirected in
3 sec
