A) \[\frac{1}{1+\log x}+c\]
B) \[\frac{x}{{{(1+\log x)}^{2}}}+c\]
C) \[\frac{x}{1+\log x}+c\]
D) \[\frac{1}{{{(1+\log x)}^{2}}}+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{\log x}{{{(1+\log x)}^{2}}}\,dx}\]. Put \[1+\log x=t\Rightarrow \frac{1}{x}dx=dt\] \[\Rightarrow dx=x\,dt={{e}^{t-1}}dt,\] then it reduces to \[\int_{{}}^{{}}{\frac{(t-1)\,{{e}^{t-1}}}{{{t}^{2}}}dt}=\int_{{}}^{{}}{{{e}^{t-1}}\left( \frac{1}{t}-\frac{1}{{{t}^{2}}} \right)\,dt}=\frac{{{e}^{t-1}}}{t}=\frac{x}{1+\log x}+c\].
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