A) \[\frac{{{a}^{x}}}{\log \,\,ae}+c\]
B) \[\frac{{{e}^{x}}}{1+{{\log }_{e}}a}+c\]
C) \[{{(ae)}^{x}}+c\]
D) \[\frac{{{(ae)}^{x}}}{{{\log }_{e}}ae}+c\]
E) \[\frac{{{a}^{x}}{{e}^{x}}}{{{\log }_{x}}a}+c\]
Correct Answer: D
Solution :
Let \[I=\int{{{e}^{x\log a.{{e}^{x}}}}}dx\] \[=\int{{{a}^{x}}.{{e}^{x}}}dx=\int{{{(ae)}^{x}}}dx\] \[=\frac{{{(ae)}^{x}}}{{{\log }_{e}}ae}+c\]
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