A) \[\frac{-\log x}{x+1}+\log x-\log \,(x+1)\]
B) \[\frac{\log x}{\left( x+1 \right)}+\log x-\log \,(x+1)\]
C) \[\frac{\log x}{x+1}-\log x-\log \,(x+1)\]
D) \[\frac{-\log x}{x+1}-\log x-\log \,(x+1)\]
Correct Answer: A
Solution :
\[\int{\frac{\log x}{{{(x+1)}^{2}}}dx=\int{\log x\,{{(x+1)}^{-2}}}}dx\] \[=\log x.\left\{ -{{(x+1)}^{-1}} \right\}\]\[-\int{\frac{1}{x}.\{-{{(x+1)}^{-1}}\}dx}\] \[=\frac{-\log x}{(x+1)}+\int{\frac{1}{x(x+1)}dx}\]\[=\frac{-\log x}{(x+1)}+\int{\left[ \frac{1}{x}-\frac{1}{x+1} \right]dx}\] \[=\frac{-\log x}{x+1}+\log x-\log (x+1)\].
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